Foundations of General Topology [1 ed.] 0125509502, 9780125509503

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FOUNDATIONS

OF

GENERAL TOPOLOGY ;t William

J. Pervin

Department of Mathematics The Pennsylvania State University University Park, Pennsylvania TT?

ACADEMIC PRESS

TEXTBOOKS IN

MATHEMATICS

Foundations of General Topology This

book

presents

a comprehensive

velopment point-set of terial pedagogically is treated,

enabling

undergraduate

topology.

oriented

to progress

of concepts

basic to modern

This is a particularly the

instructor's

point

omitted

in shorter

The student of the

proofs

The problems his mastery to additional

research.

flexible textbook of view; the

viously mtroduced to set most

background

to an understanding

tory chapter may be omitted eral the of

ma-

or upper-class

no previous

in topology

The

and rigorously

the graduate with

de-

advanced

from

introduc-

for classes

theory,

while

chapters

presev-

can

be

courses.

will find detailed as examples

explanations

for him to copy.

at the end of sections of the topics material

will test

and introduce

for further

study.

him

'^,-fy^rx^ :^M'^

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FOUNDATIONS GENERAL

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TOPOLOGY

Academic Textbooks

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JAMES SINGER. Elements of Numerical

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PESI MASANI, R.C. PATEL and D.J. PATIL. Elementary Calculus

FOUNDATIONS

OF

GENERAL TOPOLOGY

William J. Pervin Department of Mathematics The Pennsylvania State University University Park, Pennsylvania

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Preface The teaching of topology affords the instructor an opportunity not only to impart necessary mathematical content, but also to expose the student to both rigor and abstraction. This text, which grew out of lectures presented at The Pennsylvania State University, is designed to emphasize the value of careful presentations of proofs and to show the power of abstraction. Since the axiomatic method is fundamental to mathematics, the student should become acquainted with it as early as possible. The author has presented various amounts of this material to students with quite varying backgrounds; he has attempted to make this book equally flexible. The only prerequisite is the study of some analysis; even the traditional "Advanced Calculus" should be sufficient. It should be particularly noted that no part of this text requires the material sometimes labeled "Modern Algebra." Since it is possible to reach this level with no detailed knowledge of set theory, the first chapter presents the basic material in the most naive way (i.e., non-axiomatically). For students needing merely some review and practice in set theory, the second chapter introduces some important notions about cardinal and ordinal numbers, again in the naive sense. Well-prepared students may be started with the third chapter with only an occasional reference to the introductory chapters for notation. In the third through seventh chapters, the fundamentals of general topology are presented in logical order progressing from the most general case of a topological space to the restrictive case of a complete metric space. The author has found it possible to cover this material in a one-semester course with wellprepared graduate students and have time left for a few of the basic topics from the latter portion of the book (such as Sections 8.1, 8.2, 9.3, and 10.1). Including the material introduced in the problems, there is more than enough material in the entire book for a two-semester course. Students who must start with the introductory chapters will cover correspondingly less of this additional material. The proofs are given in considerable detail so that they may serve as models for the student to emulate. Only after he has had experience in giving such detailed proofs is the student prepared to omit the "obvious" steps. Although no figures are given in the text, the instructor should encourage the student to make diagrams to help his visualization. The student must be cautioned, however, that a picture is not a proof, which is the main reason for omitting them here. The

problems

at the

ends

prehension abilities and

and

of the

sections

to introduce

are both additional

to test material.

the student's Those

com-

problems

vlll

Preface

which are given with bibliographic references are usually too difficult for any but the best students and are intended more as suggestions for further reading and work. Although this material is a prerequisite for the study of many fields of mathematics, it here is presented to be studied for its own sake. It is hoped that the student finishing this text will choose to continue to study and do research in the field, and the number of references has been greatly increased for his benefit. The author has been influenced in his style of presentation by lectures by Norman Levine and the classic book by Waclaw Sierpinski. The choice of material was influenced by lectures by G. S. Young and the classic book by J. L. Kelley. The author is grateful to the many students and colleagues who have found and corrected mistakes and suggested improvements in earlier versions of this book. Throughout the book we will use the contraction "iff^" for the phrase "if and only if," and will mark the end of a proof by the symbol I. W. J. P.

Contents

Preface

Chapter

Chapter

Chapter

1 Algebra

2

of Sets

1.1

Sets

and Subsets

1.2

Operations

1.3

Relations

1.4

Mappings

10

1.5

Partial

14

Cardinal

1

on Sets

4 6

Orders

and Ordinal

Numbers

2.1

Equipotent Sets

17

2.2

Cardinal

21

2.3

Order Types

26

2.4

Ordinal

Numbers

29

2.5

Axiom

of Choice

33

Numbers

3 Topological

Spaces

Introduction

Chapter

4

Sets

36

3.1

Open

3.2

Closed

3.3 3.4

Operators and Neighborhoods Bases and Relative Topologies

Sets

Connectedness,

and and

Limit

Points

36

Closure

Compactness,

39

43 46

and Continuity

4.1 4.2

Connected Sets and Components Compact and Countably Compact Spaces

51 56

4.3

Continuous

60

4.4 4.5

Homeomorphism Arcwise Connectivity

Functions

ix

64 67

X

Chapter

Contents

5 Separation 5.1 5.2 5.3 5.4 5.5 5.6

Chapter

6

and Countability

Axioms

To-and Ti-Spaces To-Spaces and Sequences Axioms of Countability Separability and Summary Regular and Normal Spaces Completely Regular Spaces

Metric Spaces 6.1 Metric Spaces as Topological Spaces 6.2 Topological Properties 6.3 Hilbert (/,) Space 6.4 Frechet Space 6.5 Space of Continuous Functions

Chapter

Chapter

Chapter

7

8

9

69 73 79 84 87 95

Complete

99 104 108 112 115

Metric Spaces

7.1 7.2 7.3

Cauchy Sequences Completions Equivalent Conditions

118 122 125

7.4

Baire

127

Product

Theorem

Spaces

8.1

Finite

8.2

Product Invariant Properties

132

8.3

Metric

135

8.4

Tichonov Topology

138

8.5

Tichonov

142

Function

Products

129

Products

Theorem

and Quotient

Spaces

9.1 Topology of Pointwise Convergence 9.2 Topology of Compact Convergence 9.3 Quotient Topology

147 149 153

xi

Contents

Chapter

10 Metrizatlon

and Paracompactness

10.1 Urysohn's Metrization Theorem 10.2 Paracompact Spaces 10.3 Nagata-SmirnovMetrization Theorem Chapter

11 Uniform Spaces

Bibliography Index

158 160 169

11.1 Quasi Uniformization

1^^

11.2

'•^^

Uniformization

11.3 Uniform Continuity

184

11.4 Completeness and Compactness 11.5 Proximity Spaces

188 1^3

197 201

CHAPTER

Algebra of Sets

1.1

Sets

and

Subsets

It is not possible to define every term used in mathematics, but all of mathematics can be defined in terms of a few undefined concepts. The basic undefined notion with which we will be concerned is that of a set. In an attempt to clarify the intuitive notion of a set, we will note that the words "family," "collection," and "aggregate" are, and will be used as, synonyms for the word "set." Implicit in our idea of a set is the notion that an object either belongs or does not belong to any collection we call a set. The objects, or elements, which make up our sets will be referred to as points, even though they may have nothing to do with the geometric concept we have of points. Thus, if our set is the collection of all states in the United States, then we may say that Pennsylvania is one point of the set and that Europe is not a point of the set. In many discussions there is a fixed set of points from which all the sets under consideration are chosen. We will refer to this totality of all points under discussion, when such a set exists, as the universe. Our sets would then be collections of points chosen from the universe. If we choose the natural numbers (1, 2, 3, ...} as our universe, many elementary sets may be defined. The set of all even (natural) numbers, the set of all odd numbers, the set of all prime numbers, and the set of all numbers less than or equal to four are simple examples of sets.

Clearly, in any fixed universe, a set is determined by stating a property that its elements must satisfy. If P{x) is a proposition expressing a property of the point x (in the universe) which is either true or false depending on the choice of the point x, we can speak of the set of points X for which P{x) is true. If we suppress any mention of the universe, we may designate this set by {x : P{x)]. The set of all even numbers could now be written {x : x is even). The set of all numbers less than or equal to four could be written {x : x ^ 4}. If there are only a few members of a set, we may also describe it by listmg its 1

2

1. Algebra of Sets

elements. Thus, {1, 2, 3, 4} is the set of all numbers less than or equal to four. In general, we will designate points of our universe by italic lower case letters x,y,z,..., and sets chosen from our universe by italic capital letters A, B, C, ... . To designate the fundamental logical notion of membership in a set, we will write x e E and read this as "the point x is a member of the set £"," "x belongs to £"," or any other equivalent phrase which fits the grammar of the sentence. The negation of this statement, "x does not belong to E," will be written x ^ E. For example, 2 G {x : jc is even} but 6 ^ (x : x < 4}. Two sets A and B will be called equal, written A =^ B, ii and only if (iff) they contain exactly the same points. In terms of our basic notion of membership, this is equivalent to saying that A = B ifi x g A implies and is implied by x e B. Theorem

1.1.1.

For any sets A, B, and C:

[Reflexive Law]

A = A,

[Symmetric Law]

A = B implies B = A, and

[Transitive Law]

A = B and B = C imply A = C.

Proof. Each of these laws follows from a corresponding logical law. Thus, since every statement is equivalent to itself, x e A must be equivalent to x e A, so that A = A. Similarly, since x e A being equivalent X Eto B implies that x e B is equivalent to x e A, the symmetric law follows. Finally, x e A being equivalent to x e B, and x e B being equivalent to x e C, imply that x e A is equivalent to x e C, which is the transitive law. | A set J is a subset of a set B, written A ^ B or B ^ A, iff x e A implies that x e B. We will also say that B contains A, and call B a superset of A. For the negation of ^ c 5 we will write A^ B. It is important to notice that this definition does not rule out the possibility of having A = B when A ^ B. Our notion of equality shows, in fact, that A = B iff A ^ B and B ^ ^. If we wish to express the fact that A ^ B but A ^ B, we will call A a proper subset of B and write Ac B. We will warn symbol C to express both subsets following

theorem

will

the reader that many authors use the and proper subsets. The proof of the

be left to the

problems

at the

end

of the

section.

1.1.

Sets

and

Theorem

Subsets

3

1.1.2.

For any sets A, B, and C:

[Reflexive Law] [Antisymmetric Law]

A ^ A, A ^ B and B ^ A imply A = B, and

[Transitive Law]

A ^ B and B ^ C imply A ^ C.

We have already stated that we want some set to be determined a fixed universe by stating any property that its elements must If we choose P{x) to be the statement x ^ x, it is clear that P{x) for no elements of the universe. We must, therefore, accept the tionno of elements as a set. This set will be called the empty set and will always be denoted by o.

Theorem

1.1.3.

Proof. "For

For

In order

for the

all X, x g 0 implies

of this

statement,

X E £"."

This

by the be false

any

set E,

is "There

imply

definition

of the

without

causing

0 ^

x e E" would

however,

would

0 '^ E.

containment

that

that

empty

there set.

from satisfy. is true collecor null

E to he false,

have

exists exists

Since

a contradiction,

to be false. an x such

a point

the

statement

The that

negation x e 0 and

in 0 which

containment

it must

the

0 ^

be true.

is false E cannot

|

We must distinguish between the members and the subsets of a set. Thus, the set whose only member is x is {x]. Thus, 0 e {0}, and so 0 7^ {0}; that is, the set {0} is not empty because it contains something: the set 0. In general, x e £" ifif {x} c E.

Exercises

1. Let the natural

numbers

A = {x : X is even}, B = {x : there exists

{1, 2, 3, ...} be the universe.

Describe

the following

sets:

a y such that 2y = x},

C = {x: X > 1), D = {x. x = 2}, E = {x : X is even and prime), F = {x : there To

which

of these 2. Prove

exists

of these sets 1.1.2.

are

sets

equal

a y such does ? What

and that y +

the

number

subset

I = x}. 2 belong

relations

hold

? To between

which these

does sets

3 belong ?

? Which

4

1. Algebra of Sets

1.2 Operations

on Sets

The union of two sets which belong to either A and B, written A n B, is and B. The difference complement of B in A, in A which are not in B.

A and B, written A u B, is the set of all points or B or both. The intersection of two sets A the set of all points which belong to both A between two sets A and B, or the relative written A \B or Z^ B, is the set of all points

Our definitions may be rewritten as follows: A u B = {x : x e A or X G B}, A n B = {x : X G A and X e B}, and A \B = {x : XE Abut X^ B}. Some of the elementary properties of unions, intersections, and differences be mayproven immediately. Theorem

1.2.1.

[Commutative Laws] [Associative Laws] [Distributive Laws] [DeMorgan's Laws]

For any sets A, B, and C:

A uB =

B v A,

AnB=

BnA,

A v {B n C) = {A u B) An{BnC) = {A n B) A n {B o C) = {A r\ B) Au{BnC) = {A u 5)

^ C, n C, o {A n C), n (^4 u C),

C \{A u B) = {C \ A) n {C \ B), and C\{AnB) = {C\A)yj{C\ B).

Proof. Each of these laws follows immediately from the corresponding logical law. Let us prove the second DeMorgan law as an example. A point X belongs to C \{A r\ B) i& x e C and x^{A(^ B). By the logical DeMorgan law, x ^ {A r\ B) \E x ^ A or x ^ B. Using the logical distributive law, x G [C \ (^ n B)] is now equivalent to [x e C and x^A]or[xEC and x ^ B], which is equivalent to jc 6 [(C \ ^) u (C\B)], as desired. | When there is a fixed universe X under abbreviate X\E and C;^.E by C^", and call this mentE.of Other notations used for this are Notice that Zo = X and GX = 0. The Law of that CC£' = E for all sets E ^ X. With this A\B = An (C5).

consideration, we will the (absolute) compleE*^, E', r^E, and —E. Double Negation shows notation we may write

The definitions of union and intersection may be extended to any finite or infinite collection of sets as follows. If ^ is an arbitrary collection of subsets of a universe, then the union of all the sets of ^ is the set

1.2. Operations

on Sets

5

of all points in the universe which belong to at least one set in the collection

/

'2 = ' ^24 ' ...> /

/

'3 — \'^31 > '^32 % •^SS > > •''^34'

•••>

/ % i=

\^41

» •''^42> ^^43 > ^^44 > ...>

By the same procedure as in the lemma, we may list the elements in the union of the sets E^ (with possible repetitions) as follows: Xq • A direct

number. proof

In

of that

2^^ = c.

c.

Proof. By definition, 2'''> is the cardinal number of the set of all mappings of the natural numbers into the set (0, 1}. Iff is such a mapping, we may write f(n) = /„ , and thus each mapping is determined by a sequence of zeros and ones: • With each such sequence we may associate the real number S,^/„/2"; that is, the real number whose binary expansion is O./j/g/s .... Now each real number in (0, 1] has exactly one nonterminating binary expansion, while only a denumerable number of them also have a terminating binary expansion. From this it follows that 2''o = c + Nq = c. | It is clear that the same argument could be made for any finite cardinal n to show that n^^ = c. A very famous unsolved problem of set theory is the question of whether there is a cardinal number strictly between X^ and c. The continuum hypothesis asserts that there is no such cardinal number. Godel [8] has shown that if the axioms of set theory are themselves consistent, then no inconsistency will arise from the introduction of the continuum hypothesis. The generalized continuum hypothesis is that there is no cardinal number strictly between a and 2" for any infinite cardinal a.

Exercises 1. Let ^, ^ *, B, and B* be sets such that ^ nB Show that AU B ^^ A* 0 B*, and hence the defined operation. 2. Show

that

a + Xq =

a for

any

infinite

= A* n B* sum of two

cardinal

number

= 0,A ~^*,andB cardinal numbers

a.

~B* is a wel

26

2.

3. Let

a,

b, c, and

d be

cardinal

a + c < 6 -r rf, ac < M, and a'

4. Let a, 6, and c be cardinal 5. Define definition 6. Find

No! =

Find

< f.

numbers.

Does

such

that

1 %2- ... •«•...

Theorem:

and

Ordinal

a < 6 and

a < b and c < d imply

Prove that a'a"" = a'"'^,{a^y

the product of an arbitrary number of cardinal agrees with that given for a finite product.

7. Prove Konig's 8.

numbers

Cardinal

Numbers

c < ^. that

Show

that

a + c < 6 4- rf ?

= a'"^, and a^b" = {aby.

numbers

and show that your

• No-

If a; < b^^for all A, then S;i a;^ < U;^ b^ %

c« copies of a set with two elements; i.e., it can be represented by . This example shows that multiplication order oftypes is not commutative, since leu ^ co ^ w -\- oj ^ col. that, in general, a2 = a + a for any order type a.

Although multiplication of order types is not commutative, it is easy to show that it is associative and that the distributive law a(/3 + y) = a.^ + ay holds. However, the other distributive law, {a. -\- ^) y = ocy-\- jSy, does not hold! For example, (oj + 1)2 = (cd + 1) + (a> + 1) = a> + (l+a>)+l =aj + cu+l =w2+l, which is not the same order type as aj2 + 1 • 2 = 6l>2+ 2 since the second order type would be represented by a set which has an immediate predecessor to the last element, while a representative of the first order type would have no such element. It is easy to extend the above definition to the product of any arbitrary number of order types. It would also be natural to define simple powers of order types such as oj^ = o) - w = oj -\- co -\- oj -\- ...; which represented by the positive rational numbers ordered as follows:

can

be

.

Exercises

1. If f is a similarity mapping of X onto Y, then f(x) < ^(y) whenever x < y. Show that this is not true if we merely require that f be "order preserving"; i.e., that ^{x) < ^(y) whenever x < y. 2. Show

that

for

any

order

types

a, /3, and

(i) (ct + P)* = /3* + «*, (ii) (ai3)* = a*i3*, (iii) (a + iS) + y = a + (i8 + y),

y:

2.4.

Ordinal

Numbers

29

(iv) (aiS) y = ociM, (v) cx(j3+ y) = a/3 + ay. 3. Show

that:

(i) ,,* = n, and A* = A, (ii) r; + t; = Tj, but A + A =^ A, (iii) 17 + 1 + T? = % »?, and A + 1 + A = A, (iv) (v)

2.4

li m T^ n, then (1 + A) a> =

Ordinal

oj + m ^

oj + w and

OT + tu*9tM

+ co*,

1 + A.

Numbers

A partially ordered set X is wcll-ordcrcd iff every nonempty subset of X has a first element. Examples of well-ordered sets would be any finite set, and sets of order types con for any finite n. Sets of order types A, T], and a»* are not well-ordered. It is easy to see that any well-ordered set is linearly ordered. If x and y are any two elements of a well-ordered set, the subset {x, y} must contain a first element, and we would have either x < jy or j < .x. It is also clear that every subset set is itself a well-ordered set with the same ordering.

of a well-ordered

The property of being a well-ordered set is obviously possesssed by all or none of the sets of the same order type. The order type of a well-ordered set will be called an ordinal number. In the previous section we defined addition and multiplication of order types. It is an easy exercise to show that, if A and B are well-ordered sets, then {A, B} and B X A are well-ordered sets, and hence that addition and multiplication ordinal of numbers is well defined.

The following properties of well-ordered sets will be needed in our characterizations of ordinal numbers and our proof of comparability for them.

Theorem 2.4.1 . If '\ is a similarity mapping of the well-ordered set X onto the subset Y ^ X, then x < \{x) for all x e X. Proof. Suppose there were elements x such that x > f(:v). The set of all such elements, being a nonempty subset of a well-ordered set, would have a first element, say x*. If f(x*) = z, then x* > z. Since f is a similarity, and hence one-to-one, x* > z imphes that f(^*) > ^{z), that is, z > '\{z). But then z is an element smaller than x* with the same property for which x* was defined to be the smallest. | If X is an element

of the well-ordered

set X, then

the initial

segment

30

2.

of X determined y < X.

by x, denoted

Cardinal

and

Ordinal

Numbers

by X^ , is the set of all j g ^ such that

Corollary. A well-ordered set is not similar to any of its initial segments. In particular, distinct initial segments of a well-ordered set are not similar. Proof. If X were similar to its initial segment X^. , and f were the similarity mapping, then we would have t(jc) g X^ and hence f(x) < x, which violates the theorem. | We already have partially ordered order types, and so the above corollary shows that, for ordinal numbers a and j8, a < ^ ifi A is similar to an initial segment of B, where A and B are (well-ordered) sets with order types a. and jS, respectively. In order to prove that this ordering is linear, we will first consider the very powerful method of transfinite induction. We

recall

that

one way

to formulate

is as follows.

If £%is a subset

implies

n e E, then

that

necessary by

in this

1 is empty

principle

and

to any

Theorem

form

principle

of the natural

£ =

N.

well-ordered

of (finite)

numbers

Note

of the principle,

so is contained

2.4.2.

the

that since

the

N such

in E. We

that

condition

the section may

induction N,j

£ E

\ e E is. not

of N determined

now

generalize

this

set.

(Principle

of Transfinite

Induction).

// X is a

well-ordered set, and E is a subset of X with the property implies that x e E, then E = X.

that X^ ^ E

Proof, li E ^ X, then the subset of X consisting of the elements in X \ E would be nonempty, and would hence contain a first element x. Clearly every element of the section X^ , being smaller than x, is in E and so, by our hypothesis, x e E, which is a contradiction. |

Our first application of this principle will be a proof of the fact that the ordinals are linearly ordered. Theorem

2.4.3.

If a and ^ are ordinal

numbers,

then either a ^ /3 or

^ < a. Proof.

Let

respectively. wise we may U a E A and

A and

B be (well-ordered)

If either

^

associate

the

each

element

or fi is empty, first

element

of A„

has

sets the

with theorem

of B with been

order the

associated

types

is obvious. first

element

with

an

a. and

^8,

Otherof A. element

2.4.

Ordinal

Numbers

31

of B, then either every element of B has been associated with some element of A, and we must stop, or else there would be a first element of B not yet associated with an element of A, which we then associate with a. Using the principle of transfinite induction, it is clear that the process must either stop and give a similarity between B and a section oi A (^ < a), give a similarity between A and B {a = jS), or exhaust A and give a similarity between A and a section oi B {a. < j8). |

Corollary.

Every set of ordinal numbers is a well-ordered set.

Proof. Suppose 0 is a set of ordinal numbers which is not wellordered. There must then be a subset of which has no first element. Since 0 is linearly ordered by the theorem, we must be able to construct a sequence of ordinals ngN such that aj > ag > ag > ... . Let A be a set with ordinal a^ , which is then a well-ordered set. There must be initial segments A^^^ , A^ , A^ , ... whose ordinal numbers are a2 , ^3 , ^4 , ... respectively. But then subset of A which has no first element.

flg > |

^3 >

^4 >

••• gives

us

a

The previous theorem and its corollary allow us to characterize ordinal numbers in a very concise way. Since collections of ordinal numbers are well-ordered, we may speak of initial segments of ordinals. In particular, we will denote by W^i.the set of all ordinal numbers less than the ordinal number a, ordered by increasing size of the elements. Theorem

2.4.4.

Each ordinal

number a is the order type of the set W^ .

Proof. Let ^ be a set of ordinal a. For each ordinal number ^ < a, there must correspond an initial segment A^ oi A. On the other hand, each element b e A determines a segment Af^ which has ordinal j8 < a. In this way we have set up a one-to-one mapping of W^: onto A. This mapping is a similarity since a.^ < a^ implies that A^, '^ Af^ , and so 61 < ^2 , and conversely. | We may use these properties of ordinal numbers to show that the cardinal numbers are linearly ordered. In order to do this, we must make use of the well-ordering theorem proved by Zermelo [82]. Well-Ordering Theorem. Every set can be well-ordered. For a proof of this theorem, see Kamke [13, pp. 112-115] or Halmos [10, pp. 67-69]. To each ordinal number a, we may associate that of H^a . With the aid of the well-ordering

a definite theorem,

cardinal number, we may associate

32

2.

with

each

cardinal

numbers

with

ordered,

there

with

each

ordinal

that

number

a nonempty

cardinal

number.

is a unique

cardinal

set Since

first ordinal

number

Cardinal

and

composed each

set

Ordinal

Numbers

of those

ordinal

of ordinals

is well

in this set. In this way we associate

a, a unique

ordinal

a(a),

called

the

initial

of a.

Theorem or b ^ a.

2.4.5.

// a and b are cardinal

numbers, then either a ^ b

Proof. Let oc{a) and oc{b) be the initial ordinal numbers of a and b respectively. If A and B are (well-ordered) sets with these ordinals, then they have cardinals a and b, respectively. The well-ordered sets are comparable, however, so one of them is similar to a subset of the other. Similar sets are also equipotent, so one of the set A and B is equipotent to a subset of the other. Thus, either a ^ b or b ^ a. % As an example of an initial ordinal, oi is the initial ordinal of the cardinal number X^ . A particularly important property of co is that Wa, is an infinite well-ordered set, each of whose initial segments is finite. Another very important ordinal number is the first uncountable ordinal. If we take the set of all ordinal numbers with finite or denumerable cardinals, this set will be well-ordered, and hence have an ordinal number, which we denote by Q. We see that W^ is an uncountable well-ordered set, each of whose initial segments is countable. The cardinal number of Q is the second smallest infinite cardinal number and is denoted by X^ . It is clear that X^ < l^o , and the continuum hypothesis asserts that the two are equal. The crucial property of Q is found in the following theorem.

Theorem 2.4.6. // E is a countable subset of Wq , then there exists an element jS g Wq such that a < j3 for all a. e E. Proof. For each ocg E, the set {x : x < a] is countable since it is an initial segment of Wq . The union U {{x : x < a] : ocg E} is then countable by 2.1.2. | Exercises 1. A set

is well

2.

If A and

3.

Every

ordered

B are

infinite

4. A mapping/

iff it has

well-ordered well-ordered

no

sets, set

subset

show

whose

that

contains

is said to be order-preserving

order

(A,

a subset

B}

type

and

whose

iff/(x)

B

is oj*.

X A are

ordinal

well

number

< fiy) whenever

ordered. is oj.

x < j. Show

2.5.

Axiom

of Choice

that the only one-to-one is the identity mapping. 5. Show

that

6.

is the

What

number

2.5

the

initial cardinal

33 order-preserving ordinal number

mapping

of an infinite of the

set

cardinal of all

of a well-ordered has

ordinal

no immediate numbers

set onto

itself

predecessor.

of a given

cardinal

?

Axiom

of

Choice

Although we are not at all interested in presenting here an axiomatic treatment of set theory, one intuitively obvious axiom deserves mention. It is clear from the definition that when we have a nonempty set, we may choose a point from that set. The same holds for any finite number of nonempty sets — we may choose a point from each one. Strangely enough, when we have an infinite number of nonempty sets, the assumption that we can always choose a point from each set leads to some very unintuitive conclusions.

Stating the principle more precisely, if {X^];^^y^ is a collection of nonempty sets, for each A £ /I we may choose a point x^e X;^. Equivalently, we may say that there is a mapping f of A into U^g^ X;^ such that f(A) = JC;iG X; . Finally, we may state this in set-theoretic terms as follows. Axiom of Choice. The Cartesian product of a nonempty family of nonempty sets is nonempty. There are a few obviously equivalent ways of stating this principle, of which the following is historically important. Zermelo's Postulate. If ^ is a family of nonempty, disjoint sets, then there exists a set E* such that E"^ r\ E consists of exactly one point for every E e S . The interesting thing about this axiom is not the various obviously equivalent ways in which it may be stated, but the many equivalent statements of it which do not seem to have any connection with the original form. Thus, for example, the Well-Ordering Theorem mentioned the inprevious section is actually equivalent to the Axiom of Choice. Another useful statement which is equivalent to this axiom is the following, which will be used in Chapter 8. Zorn's Lemma. If a nonempty partially ordered set X is such that every hnearly ordered subset has an upper bound, then X contains a maximal element.

34

2.

Cardinal

and

Ordinal

Numbers

For a proof of this lemma assuming the Axiom of Choice and a discussion of the proof of the converse theorem, see Halmos [10, pp. 62-65]. At this point we may assume that the reader has been exposed to sufficient set theory to allow him to study profitably general topology. A few more remarks are in order, however. We note in particular that the intuitive set theory studied here also prepares the student to study some form of axiomatic set theory. The very readable book by Halmos [10] is highly recommended as a minimum. We wish to emphasize the need for further study because the intuitive set theory we have used is actually not correct. That is, our definitions will lead to contradictions. Recalling that we would allow almost any collection of objects to form a set, we should expect that we could form the set of all ordinal numbers. The following shows that this is a selfcontradictory idea. Burali-Forti

Paradox.

There

is no

set

containing

all

the

ordinal

numbers.

Proof. Suppose X is the set of all ordinal numbers. By the corollary to 2.4.3, X is well ordered, and hence its order type, say a, is an ordinal number. Thus oce X. In fact, by 2.4.4, a is the order type of W^^ = Xr, . But, by the corollary to 2.4.1, a well-ordered set cannot be similar to any of its initial segments. The contradiction lies in the fact that both X and Xoi have order type a. | The too

above

loose

paradox

with

our

is not intuitive

paradoxes

was

pointed

will

from

the

arise

paradox which

comes

from

do not contain

out

the

only

notions. Russell.

consideration

of

to form

themselves.

result

Historically,

by

trying

strange He a set

It is easy

one

showed

a set

of

we obtain

all

X which to show

that sets.

of the

first

such

a contradiction The

contains that

if we are

particular those

sets

X e X ifi X ^ X.

Since the purpose of this book is to give a careful treatment of general topology, not set theory, we will leave this topic with another recommendation the tostudent to study further in this field. Exercises 1. The

Maximal Chain Condition is: Every chain in a partially ordered set is contained a maximal in chain. Show that Zorn's Lemma, the Well-Ordering Theorem, the Maximal Chain Condition, and the requirement that every partially ordered set contain a maximal element are all equivalent.

2.5.

Axiom

of Choice

35

2. Show that Zorn's Lemma is equivalent to the following: If a nonempty partially ordered set X is such that every linearly ordered subset has a lower bound, then X contains minimal a element. 3. Use any

Zorn's

finite

Lemma number

''i , ''2 , •••, ''n ^ X =

rjACi + r^x-i +

called see

0,

a Hatncl

Goflfman

to show of

we

that

elements have

for the

[9, p. 150].

real

exists

a set B of real

Xi , JCg, ..., Jc„ G B

r^x^ + r,x^

... + TnXn for

basis

there

some

+

... + r^Xn ^

rationals

numbers.

For

r,

and

and 0;

such

for

any

rational

and

(ii)

for

some

an example

numbers

Xi&B. of the

that

numbers

every

Such

(i) for x ^t 0,

a set

use of such

B

is

a basis,

CHAPTER 3

Topological Spaces

Introduction

Before first

starting

consider

an axiomatic

the

source

of the

be based

on the generalization

We

assume

will

analysis,

and

that

treatment axioms.

and

the

has

with

pactness, connectedness and

for

topology,

Ultimately,

abstraction

reader

so is famihar

of general

notions

the

real

work

will

of the set of real numbers.

had

the

all our

we should

an

introductory

of limits,

course

continuity,

in com-

numbers.

In most modern treatments of advanced calculus it is pointed out that, although all the definitions are given in terms of the metric (distance) structure of the real numbers, everything may be rephrased in terms of the open sets alone. That is, if one knows which sets are open, one can tell which mappings are continuous, which sets are compact, which sets have limit points, and so on. In general topology we strip but even the metric structure that we know only which sets any abstract considered The subsets

set we may suppose as the open sets.

next

question

of the

collection

away not only the algebraic structure, of the real numbers, and we suppose are the open sets. More generally, for

real

of all open

to be as general possible for this

is:

that

What

numbers subsets

as possible, collection.

we know

properties

shall

be

taken

in an arbitrary and

of

so we

to

which the be

abstract shall

give

subsets collection

the

are to be of

open

axioms

for

the

set ? We

shall

try

axioms

as

as few

The most interesting fact is that we are able to define most of the important notions of analysis and prove generalized forms of many of the important theorems of analysis if we merely require that this collection be closed under arbitrary unions and finite intersections. 3.1 Open

Sets and Limit

Points

A topological space {X, S') is a set X of points and a family ^ of subsets of X which satisfies the following axioms: 36

3.1.

[0.1] (0e^)

Open

The

Sets

and

union

Limit

of any

[0.2] The intersection member of.y". {Xg:T)

Points

number

37

of members

of ^

is a member

of any finite number of members

of J^.

of oT is a

The family ^ is called a topology for X, and the members of .T are said to be the open sets of the topological space {X, ^). Two topological spaces are the same iff both the points and the family of open sets are the same in each. When the topology is clearly understood from the context, we may simply speak of the space X. A simple topology for a set X is obtained by choosing J' to be the family consisting of just 0 and X. This family satisfies the axioms [0.1] and [0.2], and hence is a topology for X. We call this topology the indiscrete topology for X. On the other hand, the family of all subsets of X satisfies the above axioms, and so yields a topology for X, which we call the discrete topology for X. It should be noted that the special statements about 0 and X in the above axioms are really unnecessary, since the union U {Gx : A g 0} = 0, and the finite intersection D {Gx : Xe 0] = X, and so 0 and X must be open for any topology. It is useful to emphasize these facts, however, since we will often show that a family of subsets of ^ is a topology for X by proving that the union of any nonempty family of open sets is open, and that the intersection of any two open sets is open. In that case, we would have to specifically check that 0 and X are open sets.

We may order the various topologies for a set X by inclusion, setting % ^\^ -^2 iff -^1 — -^2 • Although the union of two topologies for X need not be a topology, the intersection of any number of topologies is a topology, and so the family of all topologies for a set X forms a complete lattice. Clearly, the discrete and indiscrete topologies are the largest and smallest topologies for a set. Another, more complicated topology may be formed by choosing as the open sets for a set X, the empty set and all complements of finite sets. That this is a topology for X follows from DeMorgan's laws. If QGx is finite for all A, then C(U, G,) = n,(CG,) is finite, while if CG,. is finite for i = 1,2, ..., n, then 0(0^1 G^) = Uti (CG,) is also finite. Once we have chosen the open sets for a topological space (X,^), we may say that a point x is a limit or accumulation point of a subset E iff every open set containing x contains a point of E diff^erent from x; i.e., if xeGe^, then E n G\{x] ^ 0. The set of all Umit

38

3. Topological Spaces

points of a set E is called the derived set of E and is denoted by d{E). If the set X has the discrete topology, then each point is an open set and so, by considering the equation £" n {jc}\ [x] = 0, we see that the derived set of every set E is empty. On the other hand, if X has the indiscrete topology, the only open set containing a point x is X itself, and so we need only consider whether the s,&\.E r\ X \{x] = E \ [x] is empty. Clearly, the derived set of any set containing at least two points is the entire space; the derived set of a set consisting of exactly one point is the complement of that point, while the derived set of the empty set is empty. Theorem 3.1.1. // A, B, and E are subsets of the topological space {X, ^), then the derived set has the following properties:

[D.l] [D.2] [D.3] [DA]

d(0) = 0; If A ^ B, then d{A) c d{B); // X E d{E), then xed{E\ {x}); and diA u B) = d{A) u d{B).

Proof. [D.l] is true since 0 n G \ {x} = 0 for any x e X and Ge^. [D.2] follows from the fact that A ^ B implies that ^ n G\{a;} c Bn G\ {x}, and so if An G\{x} ^ 0, then also B n G \ {x} ^ 0. [D.3] is proven by considering the equation

{E \ {x})n G \ ix} = {En C{x})n G n G{x} = EnOn

G{x} = £ n G \ {x}.

To prove [D.4], we first note that d(^) u d{B) ^ d{A u B) by [D.2], since A ^ A v B and B ^ A^j B. Now suppose that x ^ d{A) u d{B), and so X ^ d(^) and x ^ d{B). From the definition, there must exist open sets G^ and Gg containing x and such that G^ n ^ \ {^} = 0 and Gb n B \ {x} — 0. Let G = G^n Gg , which is an open set by [0.2]. Then x e G, but G n A \{x] = G n B \{x} = 0, and so G n{Ayj B)\{x} = 0, which implies that x ^ d{A u B). | In our study of analysis we called a set of real numbers open iff it contained an open interval about each of its points. This family of open sets does satisfy the axioms, and so gives us a topology for the set of real numbers. This topology will be called the usual topology for the reals and will be understood to be the topology for them unless some other topology (like the discrete or indiscrete topology) is specifically

3.2.

Closed

Sets

and

Closure

39

mentioned. Of course the limit points we obtain with this topology are just those expected since our definition was just a rephrasing of the definition given in analysis. Exercises 1. List all topologies inclusion.

for a set consisting

of exactly

three

points

and

order

them

by

2. Show that the union of two topologies for a set need not be a topology for the set, but the intersection of any family of topologies for a set will be a topology for that set. Prove that the family of all topologies for a fixed set forms a complete lattice when ordered by inclusion. 3.

Let

X be any

uncountable

ments countable of

sets.

set, Show

and

that

let ^ ^

be the

is a topology

family for

consisting

of 0 and

all comple-

X.

4. Let X = N, the set of positive integers, and let ^ be the family consisting and all subsets of the form {1, 2, ..., w}. Show that ^ is a topology for X.

of 0, X,

5. Let X = N, and let ^ be the family consisting of 0, X, and all subsets of the form {n,n + \,n + 2, ...}. Show that ^ is a topology for X. Compare this topology with the topology for N formed by choosing 0 and all complements of finite sets as the open sets.

6. Let X = {a, b, c} and let d({c}) = 0, and find the derived

T = {0, {a}, {b}, {a, b}, X}. Show sets of the other subsets of X.

7. Let subsets

= {0, {a}, {b, c), X}. Find

X = {a, b, c} and of X.

let J'

the

that

derived

d({a}) = {x}, sets of all the

8. If JC is a limit point of a subset £ of a topological space {X, J'), what can be said about whether :v; is a limit point of E in the topological space {X,^*) ii ^ > ^* ? What if .:r < ^* ? 9. In the topologies

of problems

3.2

and

Closed

Sets

4 and 5, what

are the derived

sets of {1}?

Closure

The concept of a topological space has been introduced in terms of the axioms for the open sets. Let us now introduce closed sets and show that they could have been used as the fundamental notion of topology. Following our motivating example of the real numbers, we shall say that a closed set is one which contains all of its limit points. Thus, a set F is closed iff d(F) c F.

Theorem 3.2.1. If x^F, where F is a closed subset of a topological space {X, J'), then there exists an open set G such that x e G ^ GF.

40

3. Topological Spaces

Proof. Suppose no such open set exists. Then xeGg^ would imply that G^F ^ 0. Since a: ^ F, we would actually have G riF\{x] ^ 0, which means that x e d{F) by the definition of limit point. F, however, is a closed set and so d(F) s F, so that x must also belong to F. This contradiction shows that such an open set must exist. | Corollary

1.

If F is a closed set, then CF is an open set.

Proof. If X G CF, then x ^F, where F is a closed set. By the theorem, there exists an open set G-^ such that x e G^. '^ CF. But then, CF= \J{x:xeCF}^ U{G^:xgCF}c CF. Thus CF = U{G^:xeCF} which is the union of open sets, and hence an open set by [0.1]. | Corollary

2.

If CF is an open set, then F is a closed set.

Proof. If x were a limit point of F which did not belong to F, then CF would be an open set containing x which would not intersect F. But then x could not be a limit point of F. |

Corollary 3. A set is a closed subset of a topological space iff its complement is an open subset of the space. Proof. This follows immediately from the preceding corollaries. | Corollary 4. The family J^ of all closed subsets in a topological space has the following properties:

[C.l] The intersection of any number of members of ^ is a member of ^ . \Xe3F) [C.2] (0eJ^)

The union of any finite number of members of ^ is a member of ^ .

Proof. Apply DeMorgan's laws to the axioms [0.1] and [0.2]. | It is clear that any family ^ of subsets of a set X satisfying the properties and [C.l] [C.2] uniquely determines a topology for X in which the family ^ is the family of all closed sets. In particular, this topology is the family of all complements of members of J^. The closure of a set F contained in a topological space {X, ^) is the intersection of all closed subsets of X containing E. It will be denoted by c(F) or E. We see at once that c(F) is a closed set, by [C.l], and so c(F) is the smallest closed set containing E. Obviously, a set is closed iff it equals its own closure. The following is a most important characterization the closure. of

3.2.

Closed

Theorem

Sets

and

3.2.2.

Closure

41

For any set E in a topological space, c{E) = E o d{E).

Proof. Suppose x^ E^ , and so Gj n G^^-^. We must also check that Xe-T, and this is true since, by [K.l], c*(0) = 0, and so ggJ^.

(c* = c): Before we prove this part, it is important to emphasize that c(£') is the intersection of all closed sets containing E, and that, since w^e have now shown that T is a topology for X, the closed sets are just the members of the family J^. By [K.3], c*(c*(£')) = c*(£'), so c*{E)e^. By [K.2], E c c*(£). Thus c*(£) is a closed set containing andE, hence c*(£') ^ c{E). On the other hand, by [K.2], E c c{E)g.^, so c*{E) c c*(c(£')) = c{E). Thus c*(^) = c{E) for any subset E ^ X. f Exercises 1. If £ is a subset of a topological Fc X, show that £ is a closed set.

space

{X, T),

and if d(F) C £ C F for some subset

2. If (X, ^) and {X, T*) are topological spaces, what can be said about the corresponding families J^ and J^* of closed sets if -^ < .T * ? 3. What are the families of closed through 7 of Section 3.1 ? 4. Prove

the

Kuratowski

5. Show dition:

that the four

Closure

Kuratowski

sets in the topological

A, B C X.

given

in problems

3

Axioms.

Closure

A U c(^) U c(c(B)) for all subsets

spaces

Axioms

may be replaced

= c(^ U B) \

c(0)

by the single

con-

3.3. Operators

and Neighborhoods

6. Show that any closure operator

43

c has the property

that A C B implies that c(A) C c(B).

7. If .V is a point and E a subset of a topological space every open set containing x has a nonempty intersection 8. Completely define the closure through 7 of Section 3.1.

9. Show that the closure

operator

(i) U,c(£,)cc(U,£,), (ii) n^c(£:^)Dc(n;i£;i),

and

(iii) c(^) \

operators

{X, T), show with E.

in the topological

in any topological

spaces

that

x S c(£)

iff

given in problems

3

space has the following

properties:

c{B) c c{A \ B).

Give examples in which inequalities hold in each case. 10. If c and c* are the closure operators in the topological spaces respectively, how are c{E) and c*(£') related if ^ < jT*?

3.3 Operators

{X, -T) and (X, T*),

and Neighborhoods

The closure operator is just one of a number of operators which may be used to define a topology. Let us first consider the "dual" operator. The interior of a set E contained in a topological space is the union of all open sets contained in E. It will be denoted by i(£') or E. We see at once that i{E) is an open set by [0.1], and so i(£') is the largest open set contained in E. Obviously a set E is open iff £" = i(£').

Theorem 3.3.1. i{E) = Cc(C£).

For any set E in a topological space {X, ^),

Proof. If X e i{E), then i{E) is itself an open set containing x which is disjoint from C^:, and so x ^ d{GE). But x ^ GE, so x ^ (dE) u d{GE) = c(C£). Thus X e Cc(C£) and i{E) c Cc(C£). Now suppose x e Cc(C£). Immediately, x ^ c(C£'), and so x ^ {jE and x ^ d(C£'). Thus x e E, and there exists an open set G containing x such that C£' n G \ {x} = 0. Since x ^ GE, we actually have GE n G = 0, so G ^ E. We now have X E G '^ E for some open set G, and so x belongs to the union of all open sets contained in E, which is i(£'). Thus Cc(C£') ^ i-{E). I From this characterization of the interior, it is clear that if the notion of closure is chosen as the primitive one, it is possible to find the interior of sets without even finding the open sets first. Conversely, by taking complements in the previous theorem, the relation c{E) = Ci(C£') shows that if the interiors of all sets are known, the closure of any set may be found immediately. Using these relations we may obtain the following elementary properties analogous to the Kuratowski Closure Axioms.

44

3. Topological Spaces

Interior [1.1] [1.2] [1.3] [1.4]

Axioms:

i{X)==X, i{E)^E, i{i{E)) = i{E), ^nd i{A uB) = i{A) u i{B).

We will define an interior operator on a set ^ to be a mapping of ^{X) into itself which satisfies the Interior Axioms. As one might expect, a theorem analogous to 3.2.3 holds for the interior operator. Thus, an interior operator completely determines a topology (a set is open iff it equals its own interior), and in that topology, the operator is the interior. The exterior of a set E is the set of all points interior to the complement E, andof will be denoted by e{E). Thus we have e{E) = i(C£'). From this it follows that i(£') = e(C£') so that the exterior may also be chosen as the primitive notion of topology. Analogous to the Kuratowski Closure Axioms we have: Exterior

Axioms:

[E.l]

e(0) = X,

[E.2]

e(£) £ E,

[E.3] e{E) = e(Cc(^)), and [E.4]

e{A vB) = e{A) n c(5).

The

boundary

E nor

C^". It will

described refer

of a set

in the

to the

interior operator

to

be denoted literature

frontier the

E is the

set.

by b(iE'). as the

as the To

set

avoid

set

of all points Note

frontier

of all confusion,

that

interior

this

of E,

points we

is sometimes

while

some

set

which

of the will

set

to neither

not

use

the

authors are

not

frontier

at all.

From this definition we have b(£') = C(i(£') u i(C£')). Using DeMorgan's laws and the definitions of c and i, we see immediately that h{E) = h{GE) = c{E) n c(C£') = c(£') \ i{E). That the boundary may be chosen as the primitive notion of topology follows from the fact that if we know h{E), we may immediately find c{E) = Eu h{E) or i{E) = E\ b{E). Another primitive notion which may be used to define a topology is that of a neighborhood of a point, which is often more convenient than our open set notion. A neighborhood of a point is any set which

3.3. Operators

and Neighborhoods

45

contains an open set containing the point. Clearly £ is a neighborhood of ;c iff jc 6 i{E). From this it follows that a set is open iff it is a neighborhood each of of its points. In order to determine a topology from a knowledge of the neighborhoods of the points of a set, we should call those sets open which are neighborhoods of each of their points. The conditions which we must place upon the association of neighborhoods to points so that the open sets chosen in this way actually form a topology are given in the next theorem. Theorem 3.3.2. Let there be associated with each point x of a set X a collection A J.* of subsets, called neighborhoods, subject to the conditions: [N.l] Every point of X is contained in at least one neighborhood, and is contained in each of its neighborhoods.

[N.2]

The intersection of any two neighborhoods of a point is a neighborhood that of point. [N.3] Any set which contains a neighborhood of a point is itself a neighborhood thatof point. [N.4] If N is a neighborhood of a point x, then there exists a neighborhood N* of X such that N is a neighborhood of each point of A^*. Let 2r be the family of all subsets of X which are neighborhoods of each of their points; i.e., G e^ iff x e G implies that G £ ^i^*. Then ^ is a topology for X, and if ^V^is the collection of all neighborhoods of x defined by the topology ^, then jVJ^ = ^V^.for every x & X. Proof. [0.1]: Suppose G,,e^ for all A. If xeUxGx, then x e Ga for some a. By the definition of ^, Go. is a neighborhood of each of its points, so G«E.^*. By [N.3], since Ga S U; G, , U, G;iG^*. Thus U;t Ga is a neighborhood of each of its points, and hence belongs to J'. Of course oe^ since it is vacuously a neighborhood of each of its points. [0.2]: Suppose G^ and G^ belong to ^. If xe Gj n Gg, then jc e Gj and x E G2 . By the definition of S', G^ and Gg are both neighborhoods of each of their points, so Gj e ^* and GgE ^*. By [N.2], G^ n G^e^r^*, so Gi n G2 is a neighborhood of each of its points and hence is an open set. We also note that, for any x e X, x is contained in at least one neighborhood by [N.l], and this neighborhood is contained in X, so X is a neighborhood of x by [N.3]. Thus X is a neighborhood of each of its points, and so Xe^.

46

3. Topological Spaces

(^%X*, defined by setting \){x) = \{x) {or x S A and \){x) = q{x) for ac e B, is continuous. 17. A real- valued mapping f of a topological space {X, ^) is uppcr-setnicontinuous (resp. lowcr-scmicontinuous) iff the set [x •.\{x) < a} = \-\{ — ^,a)) (resp. {x : Kx) > a} = f"*((a, -r °°))) is an open subset of X for each real number a. (i) f is upper-semicontinuous (lower-semicontinuous) iff it is continuous with respect to the left-(right-) hand topology for the reals (see problem 4 in Section 3.4). (ii) f is continuous (iii) The

iflf it is both upper-semicontinuous

characteristic function continuous) E is closed iff (open).

and lower-semicontinuous.

of a set E is upper-semicontinuous

(lower-semi-

(iv) Let f be the real-valued function of a real variable defined as follows: ^{x) = 0 for X irrational or zero; ]{plq) = llq for each nonzero rational number p!q, written in lowest terms with q > 0. Show that f is upper-semicontinuous but not lower-semicontinuous.

(v) If {f;j} is a collection of upper-semicontinuous mappings, then the function f defined by setting ^(x) = ini{^^{x)} (where we allow infinite values) is also upper-semicontinuous. (vi) Iff is upper-semicontinuous on the compact space X, then f takes on its maximum value on X; that is, there is a point Xo e X such that f(jCo) > ^(x) for all x e X. (vii) Suppose X can be written as the union of the family {£„}„eN of disjoint sets, and f(^n+i) = '^n+i > '})is an open set containing X but not y. 69

70

5. Separation and Countably Axioms

Conversely, let us suppose that ^ is a T^Q-space, and that x and y are two distinct points of X. By [Tq], there exists an open set G containing of one them but not the other. Without loss of generality, let us suppose that x g G, but j ^ G. Clearly, CG is a closed set containing jy, but not X. From the definition of c({jy}) as the intersection of all closed sets containing [y], we see that y g c{{y]), but x ^ c{{y}) because of CG. Hence, c{{x]) ^ c{{y]). | A topological space ^ is a T'^-spacc iff it satisfies the following axiom of Frechet: [TJ If X and y are two distinct points of X, then there exist two open sets, one containing x but not y, and the other containing y but not X. The Ti axiom is so important that some authors (e.g. Kuratowski [17], Sierpinski [20], and Pontrjagin [19]) use the term "topological space" only when it also is satisfied. It is clear that the property of being a T'l-space is both topological and hereditary. Although every Tj-space is obviously a To-space, the converse is not true. For example, the space of problem 4 of Section 3.1 is a To-space which is not a Ti-space. In the following theorem we will give a simple characterization of T'l-spaces.

Theorem 5.1.2. A topological space X is a T^-space iff every subset consisting of exactly one point is closed. Proof. If X and y are distinct points of a space X in which subsets consisting of exactly one point are closed, then Z{x] is an open set containing y but not x, w^hile ^[y] is an open set containing x but not y. Thus ATis a Ti-space. Conversely, let us suppose that X is a T^-space, and that ;c is a point of X. By [Tj], ii y ^ X, there exists an open set Gy containing y but not x\ that is, j g Gj, c C{x}. But then C{x}= U{j : y ^ x] ^ U{Gy : y ^ x] N. Since x^—^x*, there exists an integer A^* such that x^ e G* whenever n > A^*. If m is any integer greater than both N and A^*, then x„^ must be in both G and G*, which contradicts the fact that G and G* are disjoint. |

5.2. Tg-Spaces and Sequences

77

The converse of this theorem is not true. An example of a nonHausdorflF space in which every convergent sequence has a unique Hmit was given in problem 3 of Section 3.1. A relationship between the limit points of sets and the limit points of sequences of points is given in the following theorem.

Theorem 5.2.5. // is a sequence of distinct points of a subset E of a topological space X which converges to a point x e X, then x is a limit point of the set E. Proof. If X belongs to an open set G, then there exists an integer N{G) such that a:„ g G for all n> N{G). Since the points x^ are distinct, at most one of them equals x, and so E n G \ {x} ^ 0. |

The converse of this theorem is not true, even in a Hausdorff space. Let X be the set of all ordinal numbers less than or equal to the first uncountable ordinal Q with the order topology. Although Q is obviously a limit point of its complement, no (countable) sequence of distinct ordinals can converge to Q by 2.4.6. A relationship between continuity of functions and convergent sequences of points is given in the following theorem.

Theorem 5.2.6. //f is a continuous mapping of the topological space X into the topological space X*, and A'^by the definition of a monotone decreasing base. Hence x„ —>-x and x^ —>% y, so that we have a convergent sequence without a unique Hmit. | Theorem 5.3.3. If x is a point and E a subset of a T^-space X satisfying first the axiom of countability, then x is a limit point of E iff there exists a sequence of distinct points in E converging to x. Proof. The condition is sufficient by 5.2.5. Now suppose x g d{E), and let {B^{x)] = {B^} be a monotone decreasing countable open base at X. Since x belongs to the open set B^ , the set B^n E \ {x} must be infinite by 5.1.3. By induction we may choose a point x„ in this set different from each previously chosen x^ with k < n. Clearly, x^^*- x since the sets {B^} form a monotone decreasing base at x. | Theorem 5.3.4. // f is a mapping of the first axiom space X into the topological space ^*, then f is continuous at x e X iff for every sequence of points in X converging to x we have the sequence converges to the point \{x) e X*. Proof. The condition is necessary by 5.2.6. Now suppose f is not continuous at x, so that there must exist an open set G* containing f(x) such that f(G) n CG* ^ 0 for any open set G containing x. Let {B^ be a monotone decreasing countable open base at x. Then /(5„) n CG* ^ 0 for each n, and we may pick ^„* Ef{B^) n CG*. Since ^n* ^f{Bn)i we may choose a point x^ e B^ such that /(x„) = x^*. We now have x^^ x since the sets {B^] form a monotone decreasing base at x. The sequence = cannot converge to f{x), however, since x* e CG* for all n. | A topological space {X, ^) is a second axiom space iff it satisfies the following second axiom of countability: [Cu] There exists a countable base for the topology ^. The existence of a countable base is clearly both a topological and a hereditary property. Although it is clear that every second axiom space is a first axiom space, the converse is not true. The discrete topology on any uncountable set, for example, has no countable base, since each set consisting of exactly one point must belong to any base, even though there is a countable open base at each point x, obtained by letting 5„(x) = [x]. The real number system is an example of a second axiom

5.3. Axioms of Countability

81

space since we may choose the family of all open intervals with rational endpoints as our countable base. In constructing the Lebesgue measure for subsets of the reals, it is important to have a detailed description of the structure of the open sets. The result used is that every open set in the line is the union of a countable number of disjoint open intervals. The following theorem shows that this result depends only on the fact that the real numbers form a second axiom space.

Theorem 5.3.5. In a second axiom space, every collection disjoint, open sets is countable.

of nonempty

Proof. Suppose ^ is a collection of nonempty, disjoint, open sets in the second axiom space X which has {B^} as a countable base of nonempty open sets. Since the sets [B^] form a base, for each set G in the collection ^ there must exist at least one integer n such that i5„ ^ G. Let us associate with G the smallest integer n such that B^ c G. Since the members of the collection ^ are disjoint, different integers will be associated with different members. If we order the collection ^ according to the order of the associated integers for each member, we obtain a (possibly finite) sequence which contains all the members of ^. |

The relationship between compact and countably compact sets is made clearer by application of the following theorem due to Lindelof . Indeed, it shows that the two notions are equivalent in second axiom Tj-spaces.

Theorem 5.3.6. In a second axiom space, every open covering of a subset is reducible to a countable subcovering. Proof. Suppose ^ is an open covering of the subset E of the second axiom space X which has {B^} as a countable base. Let N{^) be the countable collection of integers n such that B^ hm N{n, Gp)/n = 1, and so \J,Gx e ^. [0.2] : It is clear that Xg ^ since N{n, X) = n, and so 1 g X and Hm N{n, X)/n = 1. Suppose that Gj and Gg belong to =^. If 1 ^ Gj n Gg ,

5.4. Separability and Summary

then Gi n G2 G ^, immediately. Suppose, then, that N{n, G) + A^(w,CG) = n, it is clear that if lim N{n, CG)/n = 0. In particular, since 1 g Gj and have lim N{n, CG{)/n = 0 and lim N{n, GG2)/n = 0. equation

85

1 g Gj n Gg . Since 1 e G e ^, then 1 g Gg , we must Now consider the

N{n,Gi n G.^ + N{n, G^n CCg)+ A^(n,CG^n G^) + iV(«,CGjn CG2)= n and the inequalities

N{n,Gi n CGg)< A^(n,CGg), N{n,CGin Gg)< N{n,CgO, A^(m, CGin CG2)< iV(«,CGi). We may now calculate that lim N{n,G^n G2)/n = lim [n - A^(n,Gi n CGg)- Ar(n,CGjn G^)- N{n,CGin CGo)]/« > lim [n - N{n,CGg)- A^(n,CGi)- .^(w,CGi)]/« = 1-0-0-0=1. Hence, Gj n Gg g J^. [Tg]: Suppose x and jy are two distinct points of X. If neither point is equal to 1, then the sets {x} and {y} are two disjoint open sets containing X and y respectively. Suppose, then, that x ^ \ = y. The sets {x} and {j{x]will be the disjoint open sets desired. Non-[Ci]: Suppose {jB,J was a countable open base at 1. Since each B^ must be infinite, we may pick a point x^ g B^ such that x^ > 10", Let G = CU„{x„}. Since A^(«, G) ^ n — log^o n, we have lim N{n, G)/n^ lim (w — logiQ «)/« = 1. Hence G is an open set which contains 1. Since {B^} is a base at 1, 1 g B^ c G for some n. This, however, contradictsfact the that x^ g B^ but x^^ G. | We will now summarize the implication relationships which exist between compactness and the properties of spaces introduced in this section. Since the real numbers are second axiom, Lindelof, and separable, none of these three properties may imply compactness. Since every compact space is a Lindelof space. Fort's space shows that neither compactness nor the Lindelof property implies either separability or

86

5. Separation and Countability Axioms

the second axiom of countability. Appert's space was an example of a separable space which was not second axiom. Note that all of the above examples are Hausdorff spaces, so that none of the implications ruled out may hold even in Hausdorff spaces. By 5.3.6 and 5.4.1, every second axiom space is both Lindelof and separable. The only other impHcation not yet considered is whether every separable topological space must be a Lindelof space. Note that Appert's space is not an example of a separable, non-Lindelof space. In problems 7 and 8 we will give examples of separable Hausdorff spaces which are not Lindelof spaces. In the following diagram we denote by arrows the implications which hold in any topological space, while no other implications hold, even in a Hausdorff space. Compact Separable

- g„(x) - f„(^) > 3fl„ - a, = 2(1) (|)" = 3fl„+i .

Similarly, if x e B^ , then 6^ < Qni^) ^ 36„ while f„(x) = b^, so that 0 < g„(^) - Ux) < 36„ - 6„ = 3^»„+i.

Finally, if .jc^ ^„ u 5„ , then a„ < g^(x) < 6^ while a„ < f„(A;) < b^ , so that 3fl^+i = a„ — 6„ < g„(x) — f„(x) < ^>„— fl„ = 36„+i . Thus

the induction

is complete.

5.5. Regular and Normal Spaces

91

We now set f*(A;) = S^=if„(x) for every x e X. Since ] ]J^x) | < 6^ ,

|i;u^o| 0, B{x, e)nE\{x}

of a metric

to d.

space {X, d),

^ 0.

(ii) Xn -* X iff for every € > 0 there exists an integer whenever n > N{€); equivalently, iff d{x„ , x) -^ 0.

iV(e) such

that

x^ G B{x, e)

(iii) f is continuous at x iff for every e > 0 there exists a 8(e) > 0 such that f(B(.v, 8(e))) C B*{iix), e); equivalently, ifi x„ -> x implies that f(x„) ->%^{x). (iv) X $ c{E) iff fi(x, e) n £%= 0 for some e > 0. 7. U x„ —>% X and yn -»%>» in a metric

8. Metrizability

is a topological

space

(X, d), then : d{x,y)

< e}.

12. Let X be the set of all real-valued continuous functions defined on some fixed interval [a, b]. li d is the mapping defined by setting d(f, g) = Jo 1f(0 — 9(0 1: G c(,E) iff ^({x}, £) = 0. (iii) d{c{A), c{B)) = d{A, B). (iv) 6{A UB)

< S{A) 4- d{A, B) + h{B).

15. Show that an isometry between metric spaces is a homeomorphism. Show also that the relation of being isometric is an equivalence relation between metric spaces. 16. A To-space is the open image of a metric Ponomarev [69]).

space iff it is a first

axiom

space

(see

104

6. Metric Spaces

6.2 Topological

Properties

An uncountable set with a metric that induces the discrete topology is an example of a metric space which is neither separable, second axiom, nor Lindelof. In Section 5.4 we saw that the second axiom of countability implied, but was not implied by, separability and the Lindelof property. We will now show that these three properties are equivalent in a metric space. Theorem

6.2.1.

Every separable metric space is second axiom.

Proof. Let {X, d) be a separable metric space, and suppose E = {x^ is a countable dense subset. We assert that the family {5(x„ , 1/A) : n, ^ £ N} is a countable open base for the topology. As a countable collection countable of collections, the family is certainly countable. Since the members of the family are open (by problem 1 in Section 6.1), we must show that for each point .x:contained in an open set G, there is a member of the family containing x and contained in G. By the definition of the induced topology, there exists some e > 0 such that B{x, e) £ G, and let us choose an integer k > 2/e. Since c(£') = X, there exists an integer n such that x„ e B{x, l/k), that is, d{x, x^) < l/k. We assert that B{x^ , \jk) has the desired properties. Since d{x, x^) = d{x^ , x) < l/k, it is clear that x is contained in B{Xn , l/k). Now if y e B{Xn , l/k), so that d{y, x^) < l/k, we have d{x,y) < d{x, x^) + d{x^,y) < Ijk < e.

Thus y G B{x, e) c G, and we have B{x^ , l/k) ^ G. i

Theorem

6.2.2.

Every Lindelof metric space is second axiom.

Proof. Consider the open covering {B{x, I) : x e X] of the Lindelof metric space {X, d). By the Lindelof property, there exists a countable subcovering : {B{x^^\ 1) : « £ N}. Proceeding by induction, we consider the open covering {B{x, l/k) : x e X] and obtain a countable subcovering {B{x':l^\ 1/^) : «£N} for each positive integer k. We assert that the family [B{x^n\ l/k) : n, k e N} is a countable open base for the topology. It is certainly a countable family of open sets, so let us assume the point X is contained in the open set G. There must exist some e > 0 such that B{x, e) c G, and let us choose an integer k > 2/e. Since {5(jc^*, l/k) : « £ N} is a covering of X, there exists an integer n such that x £ B{x'^\ l/k),

6.2. Topological Properties

105

and we assert that this is the desired member of the family. Suppose yEB{x^^\ \/k), then d{x,y) < d{x, ;c(^') + ^r''^',^')

< Ijk < e.

Thus y e B{x, e) £ G, and we have B{x\!^\ Ijk) c G. | Since every compact space is Lindelof, it is clear that the three equivalent properties must all be implied by compactness. On the other hand, the example of the noncompact real line shows that the reverse implication does not hold in general. In the following diagram we denote by arrows the implications which hold in any metric space, while no other impUcations hold (compare with Section 5.4). Compact Separable o Second Axiom Lindelof

We will now obtain some additional results concerning compact subsets of metric spaces. When the concept of a compact set was first introduced, it was pointed out that the motivating Heine-Borel Theorem had no meaning in a topological space, since there was no notion of boundedness. In a metric space, however, it is easy to generalize that notion from the real numbers. We shall say that a subset £" of a metric space {X, d) is bounded iff there exists a real number /8 such that d{x, y) < ^ for all x, y e E. Since every metric space is a Hausdorff space, every compact subset of a metric space is closed. We will now show that such subsets are also bounded; in fact they have the following stronger property. If e is a positive real number, an e-nct for a subset £" of a metric space {X, d) is a finite subset F of E such that E c \j{B{x, e) : x e F}. A subset £"will be called totally bounded or precompact iff it has an e-net for every positive number e. Since an e-net is a finite set of points, it is clear that every totally bounded set is bounded. Since every compact set is countably compact, we need only show that every countably compact metric space is totally bounded in order to also prove that every compact subset of a metric space is both closed and bounded.

Theorem bounded.

6.2.3.

Every

countably

compact metric space is totally

106

6. Metric Spaces

Proof. Suppose the countably compact metric space {X, d) is not totally bounded. Then, for some fixed positive number e, X must have no e-net. Let x-^ be any point of X. Then the finite set {a;i} is not an e-net for X, and so there exists a point x^ ^ B{x^ , e); that is, d{xj^, x^) > e. Now the finite set {xj^, x^] is also not an e-net for X, and so there exists a point X3^ U,=i B{x^ , e); that is, d{Xj^, jCg)> e and d{x2 , x^) ^ e. We now proceed by induction. If we have defined a set of points {x^ , X2, ..., x„} such that d{Xi , Xj) >e whenever i^j, then this finite set is not an e-net for X, and so there exists a point x^^j^i^ ^i=i B{x^ , e); that is, d{Xi , Xj) ^ e whenever i ^ j. Now, by induction, we have defined a sequence 1/e and consider the (l/n)-net. There must be some integer i ^ N{n) such that X e B{x^'\ l/«), and so x^"-e B{x, \jn) s B{x, e), as desired. | Theorem compact.

6.2.4.

In a metric space, a set is compact iff it is countably

Proof. By 4.2.4, every compact set is countably compact, so let E be a countably compact set in a metric space. As a subspace, E is countably compact, and so, by the lemma, E is separable. By 6.2.1, E is then second axiom. Since E is also T^ by 6.1.2, the Lindelof Theorem 5.3.6 shows that E is compact (see problem 9 of Section 5.3). |

6.2. Topological Properties It

is also

easy

to

107

show

that,

in a metric

space,

compactness

and

sequential compactness are equivalent. The notion of connectedness for metric spaces will be dealt with in the problems. Exercises

1. Show that the lemma preceding 6.2.4 could totally bounded metric space is separable.

be made

stronger

by stating

that every

2. In the plane, define a new distance d* in terms of the usual metric d by setting ^*(x. y) = i^(t, e). Let /^(e) = maxi{i^(/, e)}. Then, for all k > K{€), we have dp{x^, X) < Si=i I ^i*=- a;, I + Sr=iv+i2"' < Ar(e/2iV)+ e/2 = €. I In addition to being an interesting generalization of R", Frechet space is important in that it may be considered as the topological prototype of all separable metric spaces. The next theorem shows that the study of separable metric spaces may be considered as the study of topological properties of subsets of Frechet space. Theorem 6.4.2. Every separable metric space is homeomorphic to some subset of Frechet space.

114

6. Metric Spaces

Proof. Let {X, d) be a separable metric space, and let {x„} be a denumerable dense subset of X. We define a mapping f of X into F by setting f(^) = id{x, x^), d{x, x^y ...> for each x e X. This is clearly a welldefined mapping of X onto the subset \{X) of F, and we assert that it is a homeomorphism. Suppose x and y are two distinct points of X. Since d{x, y) > 0, and the set {x^ is dense in X, there must exist an integer n such that d{x, x„) < d{x, y)/2. Hence, 2d{x, x„) < d{x,y) < d{x, x„) + d{x„ ,y),

and so d{x, x^) < d{x^ , y) = d{y, x^). But these two distances are the nth coordinates of f(x) and f(_y), so that ^(x) ^ f(jy) and f is one-to-one. To prove that f is continuous we need only show, by problem 6 of Section 6.1, that y^-^y always imphes that f(j/c) —>!(>'). But in F pointwise and coordinatewise convergence are equivalent, so we need only consider the ith coordinates; that is, does y^^-y imply that d{y^., Xi) —> d(y, x^) for each i ? But this was shown in problem 7 of Section 6.1. Finally, to show that f is open (that is, f~^ is also continuous), we will suppose that ^{y^)-> f(jy) and show that y,. —>y. Let e be an arbitrary positive number. Since the set {x^} is dense in X, there must exist an integer n such that d(y, x,^) < e/2. But ^(y/,) -^ ^(y) implies that the nth coordinates converge, so d{yk , Xn) -^ diy, x„) < e/2.

There must then exist an integer K{€) such that d{y^., x„) < e/2 whenever > kK{e). Thus we have d{yk , y) < d{y, , x,) + d{x,, , y) < e/2 + e/2 = e

whenever k > ^(e). | Exercises 1. Show

that

dp is a metric

for

F.

2. Give a direct proof of the separability subset. 3. Show that Frechet space Sierpinski [20], p. 138).

of Frechet

is homeomorphic

space by finding

to a subset

4. Let {X, d) be a fixed bounded metric space. Let J^* denote closed subsets of X and define d* by setting

d*{A,B)

= max{sup xeA

of the

y€B

xeA

Hilbert

cube

dense (see

the family of all nonempty,

inf d{x, y), sup inf d{x, y)} yeB

a countable

6.5. Space of Continuous

Functions

115

for every pair A and B in J^*. Show that d* is a metric (called the HausdorfT metric) for J-*. Prove that if X is totally bounded (compact, separable and compact), then so is J^*. For further information on this subject, see Kelley [44] and Michael [58].

6.5 Space of Continuous

Functions

In advanced analysis an important subject is the study of spaces whose points are themselves functions, the so-called "function spaces." In this section we will study one standard example of a function space, which will illustrate some of the general ideas behind this subject. The space ((£, d^ is the set (£ of all real-valued continuous functions defined on / = [0,1] with the metric d^^defined by setting rfc.(f,9) = sup{lf(0-g(0h^e/}. We

will

call

this

space

the

space

of

continuous

functions

on /.

It

is easy to see that 0 there exists a 8(e) > 0 such that d*{\{x), \{y)) < e whenever d{x, y) < S(e). As in the case of real functions, continuous functions on compact sets are uniformly continuous.

Theorem 6.5.2. // f is a continuous mapping of the compact metric space {X, d) into the metric space (^*, ^*), then f is uniformly continuous on X. Proof. Let e be an arbitrary positive number. Since f is continuous on Xy it is continuous at each point of X. Thus, for each x e X, there exists a S(e, x) > 0 such that 0 such that I f(?i) — 1(^2)I < e/5 whenever {tj^ — t^l < 8(e). We choose a positive integer n such that 1/w < 8(e), and then for each /? = 0, 1, ..., n we choose a rational number r^ such that | ^{k/n) — r,^\ < e/5. Let g be the function which takes on the value r^ at the point k/n (k = 0, 1, ..., n) and is linear in each of the intervals {{k — l)/«, k/n) for k = 1,2, ..., n. Clearly, g g (£. Now for each t e I there exists an integer k such that k/n ^ t < {k -\- \)/n. Since g is linear, | Q{k/n) - g(f) | < | Q{k/n) Q{{k+ l)/w) I, and so, by the triangle inequality.

6.5. Space of Continuous

Functions

117

I f(0 - 9(0 I < I KO - 9W«) I + I 9(^/«) - 9(0 1 < I t(0 - Qikin) I + I Qikjn) - Q{{k+ l)/«) I

< I t(0 - Kkin) I + I \{kln) - Qikjn) \ + 1Qikjn) - \{kln) I + I f(A/;;) - f((/e + l)/«) | + Im

+ l)/«) - Q{(k+ l)/«) 1

< 5(€/5) = €.

Since this inequality holds for all t e I, and the supremum of | f(0 — g(0 I is actually taken on, ^(r(f, g) < e, as desired. | Another interesting property of ((£, ^g), similar to 6.4.2, is given in the famous Banach-Mazur Theorem: Every separable metric space is isometric to some subset of (G, d^). For a proof of this theorem, the reader is referred to the text of Sierpinski [20]. Exercises 1. Show 2. that

Define the

that

d(£ is a metric

uniform uniform

convergence limit

for

d. for

of continuous

sequences mappings

of mappings

of metric

spaces

and

prove

is continuous.

3. Show that f„ 6 B(3, e) for all n in 6.5.1. 4. Suppose f is a mapping of the subset E of the metric space X into the metric space X*. The oscillation off at a point x G X is defined to be inflSC^E n G)) : G open containing x}. Prove that f is continuous at x G £ iff the oscillation of f at x is zero. Prove also that if f is uniformly continuous on E then the oscillation of f is zero at every point of X. 5. Verify that S(e) as defined continuity.

in the proof of 6.5.2 does satisfy the conditions

for uniform

6. If f is a continuous mapping of the metric space {X, d) into the metric space there exists a metric d* ior X equivalent to d such that f is uniformly continuous to d* and the metric for Y. (See Levine [52].)

Y, then relative

7. Give an example of a set which is not compact but which possesses the property that every real-valued continuous mapping of the space is uniformly continuous on the set. (For an investigation of sets which have this property, see Levine and Saunders [54] and Mrowka [60].)

8. Let S be the set of all bounded real-valued functions defined on / with the metric rfsg defined by setting f/sCf. 9) = sup{| f(0 — g(0 \:tGl}. Prove that (S, in which the distance between any two points is less than ^. Proceeding by induction, for e = 1/2 ^ we obtain a subsequence of , then we assert that x^ —>x 6 H. For each fixed integer /,

for all m, n > N{e).

If we fix n and

VSti

let m increase,

{x, - x-f

we obtain

< e,

and since this was for each /,

Thus

x„ ^-x,

if the point x does indeed

belong

to H. Now let us note

120

7. Complete

Metric Spaces

that

{Xif = {Xi— Xi« + Xi")2 = {Xi- X,")2 + (X,")2 + 2{Xi - X/") {X/") < 2(X, - X,")2 + 2(x,")2

since 2a^ < ct^ _^ ^2 Since x^ g H, S°li(^^")2 converges for every n. On the other hand, if « > A^(e),we have shown above that 2^i(x^ — ;c^^)^ N{€), 2-^ I x/^ — ^i" I /(I + I x/^ — x/ I) < 4(x,„ , x„) < e for each i. It is easy to see that, for small e, we have | x/^ — x/^ \ < 2^-^e for all m, n > N{€), so the ith coordinates form a Cauchy sequence of real numbers which must converge to some real number x^ . If we let x = (^Xi, X2 , ...>, then it is obvious that x £ F. Since we have coordinatewise convergence, x^^x by 6.4.1. | Theorem

7.1.4.

The space (£ is complete.

Proof. Let be a Cauchy sequence in G, and let e be an arbitrary positive number. There exists an integer A^(e) such that ^^(f, q) < e whenever m, n > N{€). That is, I Ut)

- fn(0 I < ML

,L) A^(e) and all tel. For each fixed t s /, this shows that N{e) and all tEl, we have — e < f„,(0 — fn(0 < + ^> and so

Ut) - ^ < Ut)

< Ut) + ^%

Letting m increase while holding n and t fixed gives us Ut) - - < KO < Ut) + ^

7.1. Cauchy Sequences

121

I WO-

KOI A^(e) and all t e I. Taking the supremum over / g / we have d^hi . t) < « for " > ^(^)- Thus f„ -> f, and we know that fG(£ by problem 2 of Section 6.5 since this convergence is uniform in the sense of analysis. | The above proof may be generalized very easily to a somewhat more interesting class of spaces. We may let (£*{X) be the set of all bounded, real-valued, continuous functions defined on the (topological) space X. The metric i(£* will be taken to be that given by setting f/e*(f, g) = sup{| f(0 — g(0 I • ^ £ ^}- It is easy to see that d^* is always a metric for ^*{X), and we could describe the space (£ as (£*(/). It is clear that the above proof shows that (£*(X) is complete for any space X.

Exercises 1. If a Cauchy

sequence

2. Every closed subset 3. Every

compact

has a convergent

of a complete

metric

space

subsequence,

metric

then

it is itself convergent.

space is complete.

is complete.

4. Show that the space (b, db) of bounded is complete.

sequences

defined

in problem

7 of Section

6.3

5. Show that the space (c, dc) of convergent is complete.

sequences

defined in problem

6 of Section

6.3

6. Show that the space (IB, d^s) of bounded is complete.

functions

defined

8 of Section

6.5

in problem

7. Let (i*{X, Y) be the set of all bounded, continuous functions of the topological space X into the metric space Y. Define a metric for (£ *iX, Y) in the obvious way, and show that if Y is complete, so is (i*{X, Y) with that metric. 8. Let X be the set of all sequences of positive integers with diixn}, 'n» = 1/" where n is the smallest integer for which x„ ^ y„. Prove that this defines a metric for X, which we call the Bairc metric. Prove that X with this metric is a complete space.

9. A mapping f of a metric space {X, d) into itself is said to be a contraction iff there exists a real number a < 1 such that d{\{x), f(>')) < 0Ld{x,y) for all x, y G X. Prove that every contraction mapping is continuous. Prove that every contraction mapping in a complete metric space has one and only one fixed point; that is, a point x such that \{x) = X. (For applications of this result, the reader is referred to the text of Kolmogorov and Fomin [16].)

122

7. Complete

10. Prove that completeness is an isometric invariant; isometric and one is complete, then so is the other.

that

Metric Spaces

is, if two metric

spaces

are

11. If f * is a uniformly continuous mapping of the subset X* of the metric space X into the complete metric space Y, then there exists a unique uniformly continuous mapping f of AT into Y such that f | X* = f *.

7.2 Completions

If a metric space {X, d) is not complete, we may wish to find a complete metric space which contains it. By using the space (£*(X) introduced in the last section, we may show directly that this is always

Theorem 7.2.1. Every metric space is isometric to a dense subset of a complete metric space.

Proof. Let us fix a point z in the metric space (X, d). We will define a mapping f of X into (£*(X) by setting, for every x E X, ^{x) equal to the real-valued function defined by setting [f(x)](^) = d{t, x) — d(t, z) for every t e X. Since ^ is a continuous function of one of its variables (see problem 7 of Section 6.1), f(^) is continuous. By the triangle inequahty we may show that | [f(x)](/) | ^ d{x, z) for all t e X, so that f(x) is bounded. Thus f(x) 6 (i*{X), and we assert that X and f(X) are isometric; that is, ^(?(f(jc), ^{y)) = d{x,y) for all x, y e X. Now 4(fG^), Ky)) = sup {| [fW] (0 - [M]

(0 1}

teX

= sup {\d{t,x)-d{t,y)\}. teX

By letting t = y, we have dii{j{x), i{y)) > d{x, y) since the supremum is not less than any element. On the other hand, suppose that, for some t G X, we had d{t, x) — d{t, y) > d{x, y).

Then, by the triangle inequality, d{t, x) > d{t, y) + d{x, y) ^ d{t, x) which

is a contradiction.

Furthermore,

if for

some

d{t, x) — d{t,y) < -d{x,y),

x e X wq had

7.2. Completions

123

then

d{y, t) < d{x,y) + d{t, x) < d{t,y),

another contradiction. Hence, for every t e X, we must have | d{t, x) — d{t, y) I < d{x, y), and so, when we take the supremum, ^^c(f(.v),f(_y)) ^ d{x, y). We now have X isometric to the subset \{X) of the complete metric space 6;*(^). The closure of f(^) in C£*(X) will be complete, and f(A^) will be dense in it. | It should be noted that the above proof depends on the completeness of (i*{X), which in turn depends on the fact that the reals are complete. In the development of the real numbers from the rationals, there is used a construction which may be generalized to any metric space and is quite interesting in itself. We will now outhne this constructive proof of the above theorem. In a metric space {X, d) we will say that two Cauchy sequences and are equivalent iff lim d{x^ , y^) = 0. One can show that this is an equivalence relation. This relation divides up the set of all Cauchy sequences in X into equivalence classes. The equivalence classes of Cauchy sequences will be taken to be the points of a new space X*. We define a metric for X* in the following way. For any two points (equivalence classes of Cauchy sequences in X) aj* and j'* in Jf*, we choose any representatives 6 jc* and (jy,,) ej* of these points and put ^*(x*,_y*) = limd'(x„ ,jv„). One can show that this limit always exists and is independent of the representatives chosen from :v* and j*. One can also show that J* is a metric for X*. We may now define a mapping f of X into Jf * which takes each point x e X into the equivalence class of all Cauchy sequences which converge to x (one representative of this class is the sequence with Xn = x for all n). One can show that f is an isometry of X onto the subset \{X) of X*. To prove that f(^) is dense in X*, we choose x* g X* and e > 0 arbitrarily. There exists an integer TVsuch that d{x,^^, x„) < e whenever m, n^ N where ex*. The point represented by the constant sequence (xj^y has the property that \im,^d{Xn, Xf^)^ e and belongs to f(X), as desired. Finally, we may show that X* is complete. Let be a Cauchy sequence in X*, and let e > 0 be arbitrary. Choose representatives e x^.* where, since these are Cauchy sequences in X and we may omit a finite number of terms from each, we may assume that 3/e such that d*{Xi*, xj*) = lim„ d{xj, xj) < e/3

124

7. Complete Metric Spaces

for all /, j > A'^(e).Thus, for sufficiently large n and all i, j > N{€), we have d{Xn^,xj) < e/3. Now, by the triangle inequality, d{Xi\ Xj') < d{x/, x„') + d{Xn\ xj) + d{x„\ x/) < 6/3 + 6/3 + 6/3 = 6

for all i,y > N{e). Thus the diagonal sequence ix/} is a Cauchy sequence in X and so determines a point jc* g X*. Clearly, d*{x^*, x*) = lim^-^(jc^'*, x/') < e, and so lim a;„* = x*. The space X* constructed above is sometimes called the complete enclosure of X. In general, a completion of a metric space X is any complete metric space which contains a dense subset to which X is, isometric. The two constructions used above give rise to the question of how many essentially different completions a metric space may have. It turns out that the completion of a space is unique up to an isometry. Theorem

7.2.2.

All completions of a metric space are isometric.

Proof. Let (X, d) be a metric space, and suppose (X*, of points of X converging to jc*. But we may also consider as a Cauchy sequence in X** and, since X** is complete, it must converge to some point :i£;**gX**. Define f(x*) = x** by this construction. One can show that this construction is independent of the particular sequence converging to x* and gives a one-to-one mapping of X* onto x**. Clearly, f(x) = x for all xeX. Now if x„^-x* in X* and x„—>-x** in X**, while yn-^y* in X* and J'n^'J** in X**, then both rf*(x*, J'*) = lim d{Xn, jy„) and ^**(x**, j**) = lim d{x^ , y^. Thus we must have d'**(f(x*), Kj*)) = ^*(x* , j'*), and so /is an isometry. | Exercises 1. Fill in the missing 2. Show that and onto. 3. The

details

the mapping

completion

of the construction f defined

of a metric

in the proof

space is separable

of the complete of 7.2.2

enclosure.

is well-defined,

iff the space

is separable.

one-to-one,

7.3. Equivalent Conditions

7.3 Equivalent

125

Conditions

In this section we will give a few conditions which are equivalent to completeness. The first result gives an application of the completion theorem of the last section. We will say that a metric space {X, d) is embedded in a metric space (X*, e^ would contradict the fact that x^ -^ x. Therefore, x g 5(x„ , e„) for all n.

126

7. Complete Metric Spaces

Now suppose {X, d) satisfies the condition of the theorem, and let be a Cauchy sequence in X. For each integer k, there is a least integer n^ such that d{Xj^^, x^) < l/l'^ for all n > n,^ . Consider the sequence of closed balls. It is clear that these are nonemptytheir and radii tend to zero. In order to use the condition of the theorem, we must show that they are nested. Nowii xeB{x^ , 1/2*), then d{x, x„ ) < 1/2*^,and so we have

d{x, x„^) < d{x, x„^^;)+ d{x„^^^, x^J < 1/2^-+ 1/2*= = l/2^-\

or X e B{x^ , 1/2*="^).By our hypothesis, there is a point x belonging to all these balls. By considering the distance d{x, x^) ^ d{x, x^) -\d{x^^, x^), it is clear that x^ -^ x, and so the sequence converges. | It is very interesting to note that even in a complete space the intersection a nested of sequence of nonempty closed balls may be empty. An example of such a space is given by Sierpinski [20]. He calls the condition of this theorem the Ascola Condition. Our final characterization complete of metric spaces leads to an interesting relationship between compact and complete spaces.

Theorem 7.3.3. A metric space is complete iff every infinite totally hounded subset has a limit point. Proof. Suppose X is complete and E is an infinite totally bounded subset of X. Since E is infinite, we may choose a sequence of distinct points from E, and then, by 7.1.1, there is a Cauchy subsequence which must converge to some point x, since X is complete. Since the sequence was composed of distinct points of E, x is 3. limit point of E. Now suppose every infinite totally bounded subset of X has a limit point, and is a Cauchy sequence in X. The set £%= U^ {a;^} is totally bounded since, for any e > 0, there exists an integer A^(e) such that d{x,ji , x^) < e whenever m, n ^ -^(^)> and so E ^ U„^i B{x^ , e); that is, the set {x„}„=i forms an e-net for E. If E is finite, one of the terms in the sequence ^2) + dAyi,y2)

is a metric for X X Y which induces the product topology. 129

130

8. Product Spaces

Proof. It is clear that ^ is a metric, so we must compare the induced topology with the product topology. Suppose E is a subset oi X X Y which is open with respect to the metric d, and let (^x,y} e E. Since 0 such that ^«^, y>, e) ^ E- Let V = B{x, e/V2) and W = B{y, e/^2), which are certainly open sets containing x and y, respectively. We assert that V X W ^ E, and this will show that E is open in the product topology. Now if {x*, y*} e V X W, then x* e V and y* g W; that is, dj,{x, x*) < e/V^ and dy{y,y*) < €j^/2. Thus d{ --i^n • By our above remarks, E^ x E^ X ... X E^ , that is, the set of all points \t,yx such that tx = Zx if A 7^ /3i , 182, ...,^n > is homeomorphic to X^^ X Xp^ X ... X Xfi^ and hence is connected by 8.2.1. This set, then, Hes completely in C. But this set contains the point (^tx^x^or which tx^ZxiiX¥^^^,^2, ..., i8„ , and /, = x, for X=^^^,^^, ..., ^„ . This point, however, lies in YlxYx , which was an arbitrary basic open set

140

8. Product Spaces

containing a , so % Y need not be either open or closed, even if Y has the quotient topology. Give examples to show that the separation and countability axioms need not be inherited by Y from X when we give Y the quotient topology.

4. If X is a countably 5. If X is completely

compact regular

Tj-space, or normal

then so is Y with the quotient in 9.3.6,

then so is X/x.

topology.

9.3. Quotient 6. Prove

that

Topology

x is closed

157

iff any

one

of the

following

conditions

is satisfied:

(i) For any equivalence class [x] in X mod r, and any open set G containing [x], there exists an open set G* containing [x] and contained in G such that G* is the union of equivalence classes in X mod r. (ii)

For any open

set G in X, the union

of all the equivalence

classes

which

are subsets

of G is open.

(iii) For any closed nonempty intersections

set F in X, the intersection with F is closed.

7. If X is closed and X is normal, closed continuous map.

of all equivalence

then so is X/t;

classes

in fact, normality

which

is preserved

have

by any

8. If X is closed and the equivalence classes in X are compact subsets, then the properties of Hausdorff, regularity, second axiom, and local compactness are all inherited by Xjx from X. 9. A mapping

f : (X/r)

—>- F is continuous

ifi ^ o x %X —>-Y is continuous.

10. If y is a quotient space of X and Z is a quotient space of Y, then Z is homeomorphic to a quotient space of X. If ti C ta , then Xjti is homeomorphic to a quotient space of X/ii. 1 1. If Xjx of X that

is a Hausdorff

X X and the

space,

x is open,

requirement

that

12. Give an example subset of X X Xsuch a nonnormal space.)

then

then Xlx

x be open

r is a closed is a Hausdorff cannot

subset

of X

space.

be omitted

of a regular Hausdorff that X/x is a Hausdorff

is a completely

this

last

which

shows

result.

by identifying all the integers. locally compact, nor first axiom.

by identifying all the points on the real but X/x is neither locally compact, nor

15. If X = [0, 1] and r = i;^ U {, }, then space

an example

subset

space X and a relation r which is a closed space which is not regular. (Hint: Consider

14. Let r be the relation on the plane obtained axis. Show that x is closed, X/x is Hausdorff, first axiom.

pseudometric

Give from

13. Let r be the relation on the real numbers obtained Show that X is closed, X/x is Hausdorff, but X/x is neither

16. Every

X X. If r is a closed

normal

X/x is homeomorphic first axiom

space.

to a circle.

CHAPTER 10

Metrization

10.1 Urysohn's

Metrization

and Paracompactness

Theorem

In the case of separable metric spaces, Urysohn [80] found necessary and sufficient conditions for metrizability. The basis for the proof of his theorem is the classical lemma mentioned in Section 5.5. We will now prove that lemma and the metrization theorem.

Urysohn's Lemma. A topological space X is normal iff for every two disjoint closed subsets F^ and F^ of X and closed interval [a, b] of reals, there exists a continuous mapping '{ : X -^ [a, b] such that \{F^ = {a} and^{F,) = {b). Proof. It is clear that if such a mapping f : X-^ [0, 1] exists, then X must be normal since t~^([0, \)) and f~^((^, 1]) would be disjoint open sets containing F^ and Fg, respectively. Consequently, let us suppose that Fi and F^ are two disjoint closed subsets of the normal space X, and [a, 6] is a closed interval of reals. Since the mapping f) defined by setting \){x) = [b — a)x + « is a continuous mapping of [0, 1] onto [a, b] (indeed, it is a homeomorphism if [a, b] is nondegenerate), we need only construct a continuous mapping g : X^[0, 1] such that q{Fj) = {0} and 9(^2) = {1}, and then f = I) o g will be the desired map. We will first define a collection {Gj. : r rational} of open sets such that c{Gj.) ^ Gg whenever r < ^ in the following way. For all r < 0 we let Gj. = 0, and for all r > 1 we let G^ = X. Next we define Gj to he X \F2 , which is an open set containing F^ with the desired property. By the characterization of normality given in 5.5.2, Gj contains an open set Gq containing F^ whose closure is also contained in G^ . Now let {r„}„gNbe a listing of all the rationals in [0, 1] with r^ = 0 and rg = 1. For each « > 3 we will inductively define the open set G^^ by taking the largest r^ and the smallest rj such that i, j < n and r^ < r^ < r^ , then using 5.5.2 to obtain the open set G^ with the property that c{G,) c G,^ and c{GJ c G,^ . We now define the desired mapping g by setting q{x) = inf{r : x e GJ. 158

10.1. Urysohn's Metrization Theorem

159

Since every xe X belongs to some G^ , but not to all of them, g(x) is a well-defined real number in [0, 1]. We note that q{x) < q iff x€ G^ for some r < q, and so {.X: Q{x) < 9} = U {G, : r < g],

which is an open set. Furthermore, r > p, and so {x : q(x) >p}=U{X\

q{x) > p iff x ^ c(Gy) for some c{Gr) :r>p},

which is again an open set. Thus Q~^{{p,q)) is always an open set, and hence g is continuous. Clearly, g(Fi) = {0} and g(i*'2)= (U- I Urysohn's Metrization is metrizable.

Theorem.

Every

second axiom

T^-space

Proof. We shall show that any second axiom Tg-space X is homeomorphic to a subset of the Hilbert cube. Let {G^},,^^ be the nonempty sets in some denumerable base for X, made infinite by repetition if necessary. Now for each integer j, there is some point x 6 Gj and, by regularity, an open set G such that x e G "^ c{G) ^ Gj . Since the collection {G„} forms a base, there must be some integer i such that X E G^ c G. Thus, for every integer j, there is an integer i such that c(Gt) c Gj . The collection of all such pairs of elements from the base is denumerable, and suppose is the nth pair in some fixed ordering. The sets c(Gj) and X \ Gj are then disjoint closed subsets of X. By problem 13 in Section 5.5, the regular second axiom space X is completely normal, and hence normal. Thus, by Urysohn's Lemma, there exists some continuous mapping f^:^^[0, 1] such that fn(c(Q)) = {0}andf„(Z\G,) = {l}. We may now define a mapping f of X into the Hilbert cube by setting f(jc) = ') = 2~" 7^ 0 since y e X \Gj . Thus, f(x) 7^ \{y), and f is one-to-one.

160

10. Metrization and Paracompactness

Now let X* be a fixed point of X, and let e be an arbitrary positive number. Let us first choose an index A^ = N{€) such that S^^^+jl^^" < e^jl. For each n such that 1 ^ w ^ A/', the mapping f„ is continuous, and so there exists a basic open set G„ containing jc* such that | \n{x*) — f„(x) | < €l\/2N whenever x e G^ %Let G = fl^^i G^ , which is then an open set containing x*. If x e G, then x g G^ for each n = \, ..., A'^,and so

^/.(tCv*),f(x))= V2-^J2-f„Cv*) - 2-"f„W]2

< vsti

11„(^*^- y^o I' + s:.;v+i 2-^"

< VN(€IV2NY+ €2/2= e. Thus

f is continuous.

Finally, suppose G is an open subset of X and y is an arbitrary point of t(G). Thus, y = ^{x) for some point x g G. As above, for some integer n, the nth pair is such that x g G^ ^ c(G^) ^ G,- ^ G. Hence f„(.\:)= 0 and f„(X \ G) = (1). Thus for any i g X \ G, d^x), f(0) ^l"** because of the diff^erence in their nth coordinates. That is, f(j£r\G)nB(f(x),2-")

= 0

Hence y g B{y, 2"^) n f(Jf) c f(G), and so f(G) is an open subset of f(X). Since f is then open, f is a homeomorphism. |

Exercises 1. If £%is an open F^ set in a normal space X, then there exists a continuous f : X ^ [0, 1] such that \{x) > 0 iff x e £; thus, X\E = \-\0). 2. A compact Hausdorff space is metrizable iff it has a countable sometimes called Urysohn's second metrization theorem.)

10.2 Paracompact Before

introducing

consider

a number

first

idea

new

base.

mapping

(This

result

is

Spaces the

topic

of basic

is a different

of paracompact

set-theoretic relation

and

on families

spaces,

we will

have

to

topological

notions.

The

of subsets

of a space

than

10.2. Paracompact Spaces inclusion.

If s^ and

161

3S are two

families

of subsets

of a set X,

then

^

is a refinement of 3S, or j/ refines 3S, iff each member of j [FN] => [PCy] =^ [PC,v] =>[PC,,] which, by the previous theorem, will prove the theorem. Suppose X is a topological space satisfying [PC,,,] and ^ is an open covering of X. By hypothesis, we may let Jt be a closed, locally finite refinement of ^ which covers X. Thus, for each H e J^ there exists a G„G "^ such that H c G^ . For each x e ^, we will set

Since M' is locally finite, U{// : x ^ H e Jf] is closed by 10.2.1, and so CU{//: x^HeJ^} = n{C//: x ^ H e JT) is open. Also, x e H e J^ for at most a finite number of //'s, so n{G^ : x e H e J^} is open, and hence V^ is an open set. Since we must either have x e H or X ^ H, it is clear that x e V^. , and so the family i^ = {V^ : x e X] is an open cover of X. We shall show that the star of i^ refines ^. Let y be any fixed point of X. We shall show that St(j, i^) is a subset of some member of ^. Since Jf covers X, we may suppose that y e H. We will show that St(3;, i^) c Gh e ^. Suppose y e Vj. for some x e X. U X ^ H, then we would have y e V^ ^ C// by the definition of Vj. , and this is a contradiction. Hence we must have x e H, and so Vj. ^ G^f since F^. is the intersection of such sets. Thus St{y, t) c G^ , as desired. The fact that [FN] implies [PCy] was proven by Stone [74], and the reader is referred to his paper for a long but straightforward proof of this result. Since each a-discrete system is a-locally finite, the next implication is obvious. Finally, suppose X is a topological space satisfying [PCjy] and ^ is an open covering of X. By hypothesis, there exists a a-locally finite open cover Jf which refines ^, which we may write as Jf = ^^ne^-^n with each J^^ locally finite. Now for each H e Jf , H e Jf „ for some smallest «, and we may let F^ = // \ U{//* : W e :^ ^ for i < n]. Since V^ ^ H, the family f = {Vj^ : H e J^] refines Jf and so also ^. For each X E X, there is a smallest integer n such that there exists an H^. e Jf „ containing x. Since x e V^ , 'f covers X. We note that H^. is disjoint from each V^ for H e J^^ with i > n (where we choose the smallest possible index i). Lastly, since each Jf^ is locally finite, for each i = \, 2, ..., n there exists an open set W^ containing x which intersects at most a finite number of members of Jf^ . If we let W = H^n W^n ... n W^ , then W is an open set containing x which intersects at most a finite number of members of 'f, so that t" is locally finite. |

10.2. Paracompact

Spaces

167

The characterization [PCy] allows us to state some simple sufficient conditions for paracompactness. We say that a topological space is CT-compact iff it is the union of a countable collection of compact sets.

Corollary. Every regular Lindelof space is paracompact. Every regular space which is either second axiom or a-compact is paracompact. Proof. Every open covering of a Lindelof space has a countable subcovering, and so, automatically, a a-discrete refinement. Every a-compact and every second axiom space is a Lindelof space. | The behavior of paracompact spaces with respect to products is of great interest today. The space of problem 8 in Section 5.4 due to Sorgenfrey [73], is an example of a regular Lindelof space whose product with itself is not normal (problem 3 in Section 8.2), and so the product of paracompact spaces need not be paracompact. Dieudonne [36] showed that the product of a paracompact space with a compact space is paracompact. This result was strengthened by Michael [56] in the following theorem. Theorem 10.2.6. // X is a regular paracompact a-compact space, then X x Y is paracompact.

space

and

Y is a

Proof. Let Y = U^g^^n with each y„ compact, and suppose that ^ is an open cover oi X X Y. For each point z = in ^ X Y, there are open sets V^ and W^ in X and Y, respectively, such that 0 iff x g B^,;, . For each fixed integer «, the family {B,^ %: Ae A^} is locally finite, and so, for each fixed point x g X, hi.M ^ 0 for ^t most a finite number of values of A. Hence 1 -f- S^f^,^(.x;) is a well-defined continuous mapping of X which is never less than one. From this it follows that we may define a continuous mapping g„., :Z^[0,1] by setting r

..-H

1-1/2

Again we see that %nA^) > 0 ifT x g B^j^ , while, for a fixed integer n and fixed point xg X, g„,;i(x) ^ 0 for at most a finite number of values of A. It is obvious that ^x^n a(^) < 1 ^^id it is easy to verify that S,[9n.A(^) - 0, while \)ri.?iy) = 0, so \{x) ^ \{y), and f is one-to-one. Now suppose that xg X and e > 0 are given. First choose an integer A^ = iV(e) such that 2"^ < 6^/4. By the local finiteness property, there

172

10. Metrization and Paracompactness

must exist an open set G containing x which has a nonempty intersectionatwith most a finite number of the sets B^ ^ with n ^ N. Let us denote these sets by -S„. ;,_, where n^ ^ N for i = 1, ..., K. Since each function l)^.^ is continuous, we may find, for each e A, an open set Gn_x containing x such that 1f)n.A(.v) - Kjy) for

every

y e Gnx • Let

I < e/V2Z

us set

which is an open set containing x. We now note that for (n, X) e A but not equal to some (n^ , A^), ^nA^) = ^nAy) — ^ ^o^"every y £ G*. Thus we have, for jy e G*, nN

X

< 2 2) 2-" = 2(2-^) < 2(e2/4) = £2/2.

Thus

we

have

shown

V(f(^l

that

f(>')) = \/S„.A [i)nM - i)Uy)Y < ^

for all y e G*, and so f is continuous.

Finally, let G be an arbitrary open set in X, and choose a point x e G. We must have x e Bnj_ ^ G for some (n, A> e /I. Let ^n.x{^)ywhich is a positive real number, be denoted by 8. If \{y) is a point of f(X) such that dH{'^{^)My)) < ^> then I)n.;i(>')is also positive, and so jy e B^^x^ G. Thus f-i(B(f(^).S))^G so that f is open.

From

this it follows

that f is a homeomorphism.

|

By a construction very similar to that used to prove Stone's theorem 10.2.2, it is possible to show that for every open covering of a metric

10.3." Nagata-Smirnov Metrization Theorem

173

space, there is a a-discrete open cover which refines it. Hence, every metric space is a Tg-space with a a-discrete base. Since every a-discrete family is a cr-locally finite family, we may state the following: Bing Metrization Theorem. A topological space is tnetrizable iff it is a T^-space with a a-discrete base. If the condition that the topological space be a r^-space is removed, the above theorems yield pseudometrizable spaces (see 9.3.7). Examples may be given, however, to show how the condition of regularity is essential. Exercises 1. Show

that

dH^

is a metric

2. Show that Urysohn's

for

metrization

W.

theorem

3. Show that Urysohn's second metrization corollary to Bing's metrization theorem. 4. Give an example of a second axiom not metrizable (see Smirnov [71]).

is a corollary theorem

Hausdorff

space

to Bing's metrization

(problem which

2 in Section is not regular,

theorem. 10.1) is a and hence

5. A topological space is called locally metrizable iflf every point is contained in an open set which is metrizable. Prove: If a normal space has a locally finite covering by metrizable subsets, then the entire space is metrizable. Corollary: A locally metrizable Hausdorff space is metrizable iff it is paracompact (see Smirnov [71]).

CHAPTER I I

Uniform Spaces

11.1

Quasi

Uniformization

We have already discussed a few nontopological notions, such as uniform continuity, uniform convergence, and completeness. Although these concepts are valuable in the study of metric spaces, they are lost in the generalization to topological spaces. We w^illnow^consider another generalization of metric spaces in which these concepts are meaningful. A quasi-uniform space {X, U) is a set X together with a family U of subsets of X X X which satisfies the following axioms: [U.l]

For

every

[U.2]

If D 2 u G H, then

[U.3]

If u 6 U and

[U.4]

For

every

u e U, i^ ^ "• d e U.

d e U, then

u e U, there

u n o e U. exists

a d g U such

that

d o o c

u.

The family U is said to be a quasi uniformity for X. The quasiuniform space {X, U) will always be considered to be a topological space, with the topology obtained by using as the family of all neighborhoods a pointof x e X the sets of the form u{x) with u g U. That the family of sets u{x) may be chosen to be the neighborhoods of the point x follows from the fact that the above axioms are just uniform forms of the usual neighborhood axioms [N.l] through [N.4] of Section 3.3 (see [67]). As generalizations of metric spaces, it should be possible to obtain a quasi-uniform space from any metric space in some natural way. Suppose ^ is a quasi metric for X; that is, a mapping of X X X into the nonnegative reals satisfying the axioms [M.l] and [M.2] for metrics (see Sections 6.1 and 9.3). The family U of all sets u which contain a set of the form w, = {(x, y} : d{x, y) < e}

for some e > 0 is a quasi uniformity for X. We call this quasi uniformity the metric quasi uniformity induced by d. 174

11.1.

Quasi

Uniformization

175

Since, by [U.2] and [U.3], a quasi uniformity is a filter (see Section 8.5) of sets containing t;^.(the diagonal) by [U.l],we may discuss bases and subbases for a quasi uniformity (see problem 2 in Section 8.5). Thus the e-neighborhoods of the diagonal, which we have called w, , form a base for the metric quasi uniformity. As one might expect, not every family of sets S containing the diagonal is a subbase for a quasi uniformity, for it is required that for every u e S there exist a De 3 such that d o d ^ u. As with topological spaces, it would be useful to know when, for a given quasi-uniform space {X, U), there is a quasi metric for X which induces U; that is, under what conditions is a quasi-uniform space quasi-mctrizable. The result is much simpler than that for topological spaces, as the following theorem shows. Theorem 11.1.1. A quasi-uniform space is quasi-metrizable quasi-uniformity has a countable base.

iff its

Proof. Clearly, if ff is a quasi metric for X, then the family of sets of the form °'i/n = { '• d{x,y) < 1/n},

for each n eN, is a countable base for the metric quasi-uniformity. Now suppose that {X, It) is a quasi-uniform space with the countable base {D„}„eNfor its quasi-uniformity. We may define, by induction, a family of sets {u„}„gNwhich forms a base for U and is such that u„ ^ d^ and u,^ o u„ o u„ £ u^_i for every n eN. We let Uj = Dj , and suppose that u,^_l has been defined. By [U.4] there exists a set u„* 6 U such that u„* ou„* c u^i J and then there must also exist a set u** 6 U such that u** o u** c u^*. We will let u,^ = u** n d^ so that u„ £ d^ immediately.i;^ Since ^ u^ , by [U.l], we have U„ OU„ OU^ = c

'^X°^n

OU« 0»n

u„ o u„ o u„ o u„

c (u** n D^)o (u** n dJ o (u** n dJ o (u** n dJ c (u** o u**) o (u** o u**) c u^* o u^* c u^_^ ,

Next, we define a real-valued

mapping cp on X X X by setting

cp{x,y) = inf„{2-": gu„}

176

11. Uniform Spaces

for each {x, y} e X X X. Finally, we define the desired quasi-metric by letting

d

d{x, y) = inf 2) + 1) 2- we may

now

calculate

Un O D„2-" = Un O Dp2-("-i)+2^ = U„ O U„ O Dj,2- cp{x) > (m - 1)2-^ and m*!'" > cp{y) > (w* - 1)2"^ By the definition of so that i3> : a > t and

p > t} for each

ordinal

t < ii. Show

that

U is not

metrizable.

3. Show that there is only one uniformity for the set of all ordinals less than the first uncountable ordinal that induces the order topology (see Dieudonne [35]). For information concerningconditions under which a space has a unique uniformity, see Doss [37].

184

11. Uniform Spaces

4. Tukey [24] gave the following derivation of uniform spaces. Let T be a family of open covers of the space X satisfying the following properties: (1) If a cover 'S E: F, then any refinement of 'S belongs to F; and (2) Any two members of F have a common star refinement in F. Then the family {G X G : G e ^} for ^ 6 T is a base for a uniformity for X. U The neighborhoods of a point x in the induced topology are the members of the family {St(jc, ^) : ^ e T}. In the uniformity U, the family F has the property that each member ^ of T is a uniform cover of X\ that is, there exists a set u G U such that {\\{x) : x^X) refines ^ . Conversely, the family F of all uniform covers of a uniform space satisfies the above properties (1) and (2).

11.3 Uniform

Continuity

The first property of metric spaces which we shall introduce in the more general setting of uniform spaces is that of uniform continuity of functions. Recalling that it involves finding a distance which satisfies certain requirements for all points of the space simultaneously, we see that it is natural to consider this notion in uniform spaces since each member of a uniformity gives a particular neighborhood of each point of the space simultaneously. A mapping 93 of a uniform space (X, U) into a uniform space (Y , 93) is uniformly continuous iff for every D G SB there exists a u e U such that i^fix^, (p{x^^ g d whenever e u. For a mapping 99: {X, dx) —> {Y, dy) of metric spaces, it is clear that uniform continuity in the metric uniformities is equivalent to the requirement that for every e > 0 there exists a S > 0 such that dY{ F X Y, which we associated with the mapping fp : X ^ Y by setting 992((^i >^2)) = (9'('^i)> 9'('''^2)) i^i problem 8 of Section 8.1. With this notation, 9? is uniformly continuous iff for every D G 93 there exists a u g U such that (p,< vi , yi)} : e u} or of the form {«xi , X2>,(yi , y2>y : ^2) I < ^-

11.3. Uniform Continuity

187

If we now let x^^= x and Xg = jj = jg = J. we have (x, y} e u implies that d{x, y) < e. Thus u c to^ , so to^ 6 U for all e > 0.

Conversely, if to, g U for every e > 0, then for any fixed e > 0, a),/2 e U. Now if (x-^^, x^) e w.^^ so that d{xi , x^) < e/2, and GD. Furthermore, yeF, so that for any point x eF, c u, which means that x g u{z). Thus F c u{z), which implies that u{z) e ^, and J^ must converge to z. | Perhaps a more important question about Cauchy filters is whether they are uniform invariants. This will follow from the following result.

Theorem 11.4.2. // 9 : {X, U) -^ (7, 53) is uniformly continuous and #" is a Cauchy filter {or the base of a Cauchy filter), then (p{-^) = {(p{F): F E J^} is the base of a Cauchy filter.

190

11. Uniform Spaces

Proof. It is clear that (p{'^) is the base of some filter in Y, so suppose r»e 33. There must exist some set u g U such that (p2{u)^ o, since (p is uniformly continuous, and there must exist some set F e ^ such that F X F ^ u, since .#" is a Cauchy filter (or the base of a Cauchy filter). We assert that (p{F) x (p{F) c o, which will show that